Sums and Products 2018900205, 0999342819, 9780999342817

Table of contents :
Introduction
1 Telescoping Sums and Products in Algebra
Telescoping Sums and Products in Trigonometry
Complex Numbers and de Moivre’s Formula
The Abel Summation Formula
Mathematical Induction
Combinatorial Identities and Generating Functions
Sums and Products in Number Theory
Problems
1 Easy Problems ....................
2 Medium Problems ..................
3 Hard Problems ....................
Solutions
1 Solutions to Easy Problems .............
2 Solutions to Medium Problems ...........
3 Solutions to Hard Problems .............
References

Citation preview

Sums and Products

Sums and Products

Titu Andreescu

Marian Tetiva

Library of Congress Control Number: 2018900205

ISBN-10: 0—9993428—1-9 ISBN-13: 978-0-9993428-1-7 © 2018 XYZ Press, LLC All rights reserved. This work may not be translated or copied in whole or in part without the written permission of the publisher (XYZ Press, LLC, 3425 Neiman Rd., Plano, TX 75025, USA) and the authors except for brief excerpts in connection with reviews or scholarly analysis. Use in connection with any form of information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed is forbidden. The use in this publication of tradenames, trademarks, service marks and similar terms, even if they are not identified as such, is not to be taken as an expression of opinion as to whether or not they are subject to proprietary rights. 9 8 7 6 5 4 3 2 1 www.awesomemath.org Cover design by Iury Ulzutuev

Contents

Introduction

1

Telescoping Sums and Products in Algebra

19

Telescoping Sums and Products in Trigonometry

53

Complex Numbers and de Moivre’s Formula

93

The Abel Summation Formula

107

Mathematical Induction

121

Combinatorial Identities and Generating Functions

139

Sums and Products in Number Theory

159

Problems

171 171 177 184

1

Easy Problems ....................

2

Medium Problems

3

Hard Problems ....................

..................

193 Solutions 1 Solutions to Easy Problems ............. ....... 193 231 2 Solutions to Medium Problems ........... 279 3 Solutions to Hard Problems .............

vi

Contents

References

335

Other Books from XYZ Press

337

Introduction Let us start with a simple (?) question. In a chess tournament there are n players who play eliminatory games until only one winner remains. Thus, in the first round, the players are arbitrarily playing games (chosen by drawing lots), and only the winners of these games go to the second round. If the number of players is odd there is one player staying aside, but he/she (and all winners) will take part to the drawing for the second round. The process repeats in the second and all the following rounds until, as we said, finally,

only one winner remains (and he is declared the winner of the championship). The question is: how many games are necessary in order to establish the champion? Well, you might need a moment of thinking, and we strongly advise you to take it (or, maybe, you already got the answer, which is great). You will immediately see that if n = 2'" is a power of 2, then in the first round there are 2m_1 games (and 2m_1 winners from these games accede to the second round), and the process goes on and on so that there will be 2m_j games in the jth round. The total number of games will then be

2m—1+2m-2+...+2+1=2m—1=n—1. Although we cannot quite use this reasoning in the general case, the answer n — 1 is correct for each and every value of n, because in each game precisely one player is eliminated and, in order to arrive to the situation when only one player still stands, n — 1 players must be eliminated, so, 77. — 1 games are needed to see who the champion is. The problem is solved, but we won’t stop here. This is because the reasoning in the particular case of n = 2m furnishes a hint for the general case. Namely,

2

Introduction

in the first round the number of games is k if n = 2k is an even number, and it is also k if n = 2k: + 1 is odd (and a player is forced — also by drawing, to

make the competition fair — to stay aside). This number can be expressed (for both cases of even and odd n) as [11/2] where [x] denotes the integer part of x (or the floor function of ac) — the largest integer which is not greater than a: (that is, [as] = p if and only if p is the only integer such that p _ 1 is given: it ensures |1/a| < 1, hence 1 1-,. lim (—) = 0 71—)00

a,

Introduction

17

whenever (x0121 is a sequence of real numbers with limit 00. In particular oo I I

’°=1

1

1 + —k

a2

]_ =

lim

n—)oo

——1

— 2n+1 0’

1

1

=

1

“—

1

=

1

* 25

a

2

.

a2 — 1

We end this introductory part with a question related to the problem from which we started. For

a: f(w) = m — [5]

and nk defined by m = n (an arbitrary positive integer) and nk = f (nk_1) for k 2 2 we have seen that

l—Jl—J Can we compute this sum by telescoping? (The answer is yes. So, find how.)

Chapter 1

Telescoping Sums and Products in Algebra One of the most useful techniques for computing sums and products is the use of the identity

(a2—a1)+(as—a2)+---+(an+1 —an) =an+1-a1, valid for any complex numbers a1, . . . , an. Therefore, if we need to compute a TL

sum 2 bk, we might try to find numbers (11,. . . , an“ such that k=1 b1=a2—a1, b2=a3—a.2,...,bn=an+1—an

and then apply the previous identity to deduce that the sum we are looking for is simply an+1 — a1. If we can do that, we say that the sum is telescopic,

or that it telescopes (as shown in the introduction). Finding the numbers a1, . . . ,an is the hard part of the game and lots of practice is certainly helpful! Note that it is always possible to find (11, . . . ,an+1 as above, namely choose a1 = 0, then 0.2 = b1, a3 = b1 + b2, . . ., an+1 = b1 + ~ ~ ~ + bn. Of course, this is not very satisfying for our needs... Let us start with a few classical examples. You certainly know the following

20

Chapter 1. Telescoping Sums and Products in Algebra

identities

in: k = n(n + 1), k=1

2

£19 2 n(n+ 1)(2n + 1), k=1

6

a. ._|||

£193 2 [71,012+ 1)]2.

'n.

n.

'n,

What about 2 k4, or 2 1:5, or more generally 2: H" with N a given positive

k=1

k=1

k=1

integer? It turns out that one can actually find similar formulae for these sums, but the formulae become fairly complicated when N is large. The key idea is the use of the binomial theorem in the form

(Ic+1)N+1 —kN+1 = (N:1)kN+ (N:1)kN‘1+-~+ (N;1>k+1. Thus, adding these relations for k = 1,2,... ,n yields (you see the telescope,

don’t you?)

(n+1)N+1—1=(

N+1 )(1N+2N+---+nN) 1

N 1 +( J)(1N—1+2N—1+---+nN—1)+---

+( N1:; 1 )(1+2+---+n)+n. This shows that if we can compute lj +2j + ~--+nj for j = 1,2,...,N — 1, then we can also compute it for j = N. For instance, take N = 1, then the previous identity becomes

(n+1)2—1=2(1+2+---+n)+n,

Chapter 1. Telescoping Sums and Products in Algebra

21

and a simple algebraic manipulation shows that this recovers the classical formula 1+2+---+n=

n(n + 1) 2

Next, take N = 2, then we obtain

(n+1)3—1=3(12+---+n2)+3(1+2+~-+n)+n, in other words

12+---+n2=

(n+1)3—1—3(1+2+---+n)—

3 n3+3n2+3n—

3n(n+1) T—

3 _ 2n3+6n2+6n—3n2—3n—2n

_

6

_ 2n3+3n2+n _ n(2n2+3n+1) _ 6 _ _ n(n+1)(2n+1)

6

_—é———.

Example 1.1. Prove that

n

_ n(n + 1)(2n + 1)(3n2 + 3n — 1)

30

2'94 _

'

Solution. From the binomial theorem we have

(k + 1)5 — k5 = 51:4 + 101:3 + 101:2 + 5k + 1, thus it follows that

52k4+102k3+102k2+52k+21—Z((k+1)5—)=k5 —(—n+1) k= 1

k:1

Hence

5 :1 k4: (n+1)5— 10

712(714—1—1)2 _10.n(n + 1)6(2n + 1) _5.n(n2+ 1)

—(n+1),

22

Chapter 1. Telescoping Sums and Products in Algebra

yielding

"

n(n + 1)(2n + 1)(3n2 + 3n — 1)

2 k4:

30

Example 1.2. (IMO Longlist 1977) Evaluate 'n,

Zk(k+1)---(k+p— 1), where n and p are positive integers. Solution. We have

ik(k+1)---(k+p—1)=ik(k+1)'”(k+p)_(k_1)k"'(k+p_1) k=1

p+1

_n(n+1)---(n+p) ' p+1 For example

Zk(k+1)=n(n+1)(n+2) (as we already have seen in the introduction) and n

2,909+ 1)(k+2) = n(n+ 1)(n4+ 2)(n+3). k=1 Example 1.3. For positive integers n and p Z 2 evaluate

;k(k+1)-- -1(k+p— 1) Solution. We have

Z::llc(k+1)-- ~1)=(k+p—1

(k+p—1) p—1:::1Mk+1)-- -(k+p—1)

Chapter 1. Telescoping Sums and Products in Algebra n

1

23

1

=p—1k2:41(k(k+1)m(k+p—2) _ (k+1)---(k+p—1)) _ 1 < —p—1

1

_

1

(p—1)!

For instance,

)

(n+1)---(n+p—1)

'

i— 1 1___1_ k—__1(k+1) n+1 1

1

1

Zk——(k+1)(k+2)= 5 (5— (n+1)(n+2))' Example 1.4. (IMO Longlist 1970) For even positive integer n, prove that

”

_

1) :3 1:1

1

i+1 . — =

i

"/2

1

n + 2i" 2 Z: 1:1

Solution. Induction on n works. It is easy to verify the case n = 2, so assume the result is true for some even natural number n. Then n+2

n

. 1 1 1 1 _1l+1._=___ _1t+1._ ;( ) 2' n+1 n+2+;( ) i

and by the inductive assumption, this will be equal to

1 1 2 1 + (— ————+—+2 n+1 n+2 n+2 +4 Since

1 )

+n+n

-

L+;—2( 1 + 1 ) n+1 n+2_ (n+2)+n (n+2)+(n+2) m

1 the sum becomes 2 Z — which finishes the proof. 71+ 22' The next example involves factorials.

24

Chapter 1. Telescoping Sums and Products in Algebra

Example 1.5. Evaluate

12-2!+22.3!+---+n2(n+1)! Solution. We try to find numbers a), such that k2(k+ 1)! = ak+1 — ak for 1 S k g n. It is natural to look for ak of the form ak = klbk for some number bk. Indeed, the previous relation becomes then the much easier

1:209 + 1) = (k + 1):”,+1 — bk. Next, we try to find bk of the form bk = P(k) for some polynomial P. Thus we ask that

k2(k + 1) = (k + 1)P(k + 1) — P(k). If we are lucky, the previous equality holds for all integers k and so, by con— sidering degrees P must be quadratic and monic, say

P(X) = X2 + cX + d. Plugging in the previous relation yields

k2(k+1) = (k+1)((k+1)2+c(k+1)+d) —k2—ck—d. Identifying coeflicients easily yields c = —1 and d = —2. Thus

i k2(k + 1)! = flag + 1)!((k + 1)2 — (k + 1) — 2) — k!(k2 — k — 2)] k=1

k=1

= (n+ 2)(n — 1)(n + 1)! — (1 + 1)(1 — 2)1! = (n— 1)(n+2)! +2.

This example illustrates an extremely important technique for finding telescoping sums and products. We did not immediately see the correct formula, so instead we guessed What the form of the answer might look like. Since we weren’t sure of the exact form, we guessed a form with some undetermined parameters (in this case, the degree of P and its coeflicients). We then used the fact that we wanted the sum to telescope to let us solve for these parameters. Since we guessed the correct form, we were successful and we solved the problem.

Chapter 1. Telescoping Sums and Products in Algebra

25

Example 1.6. Evaluate n

219K192 + k + 1). k=1

Solution. We try to write

k!(k2 + k + 1) = ak+1 — ak for some numbers ak. To get rid of kl, let us choose ak = klbk for some numbers bk that we still have to find. Then

ak+1 - ak = (k + 1)!bk+1 — klbk = k!((k + 1)bk+1 - bk), so we need

(k+1)bk+1 —bk =k2+k+1. But there is an obvious choice: set bk = k for all k. Thus we can take ak = 19-19! and the desired sum telescopes nicely: n

n

Zk!(k2 + k + 1) = XKk + 1)!(k + 1) — klk] = (n + 1)!(n + 1) — 1. 19:1

—

Example 1.7. Evaluate 2— 3192 — 1

2—(k3

’02

Solution. First, observe that 1

1

(k3 — k)2 = 19209 — 1)2 ‘ 1:209 + 1)? Now the series telescopes as

26

Chapter 1. Telescoping Sums and Products in Algebra

Example 1.8. Compute

1 - 3! _

2 - 4!

3

+2 ! ...+ m4,

32

3”

Solution. Observe that

1909 + 2)! _ (k + 3 — 3)(k+ 2)! _ (k + 3)(lc + 2)! — 3(lc + 2)! 3k

3k 3k _ (k+3)l—3(k+2)l (k+3)l (k+2)! _ 3k = 3k _ 3k—1 '

We obtain therefore a telescoping sum

ik(k+2)! =i((k+3)! _ (k+2)!) = (n+3)! _3_!= (n+3)! _6.

k=1

3k

k=1

3k

316—1

3n

30

3n

Example 1.9. Evaluate

2”: k

4k+\/4k2—1

= 1 \/2k+1+\/2k—1'

Solution. This problem requires some algebraic skills. To simplify the formulae let us denote

a=v2k+1,

b=v2k—1.

Then 2k + 1 = a2 and 2k — 1 = b2, so that 4k = a2 + b2 and 4k:2 — 1 = a2b2. Thus a3—b3

4k+\/4k2—1

_a2+b2+ab_a2+ab+b2_ a_b _a3—b3

\/2k+1+\/2k—1_

a+b

_

a+b

_a+b _a2—b2'

Since

a2—b2=(2k+1)—(2k—1)=2, weobtain

4k+\/4k2—1 _ a3—b3 _ ‘/(2k+1)3— ‘/(2k— 1)3 ' 2 _ 2 x/2k+1+\/2k—1_

Chapter 1. Telescoping Sums and Products in Algebra

27

Telescoping, we find

5:

4k+\/4k2 — 1

_ w/(2n+1)3 — 1

'

2

16:1 x/Zk + 1 + x/2k :1' _ Example 1.10. Evaluate the sum 2

22

23

2n+1

3T1+m+m+m+32n—+1 Solution. We know that

1_a—1_

1+1

a+1_a2—1_ which implies

1

(12—1

1

_

1+1

2

a+1

1

a—1

’

1

2 a+1_2(a-1)

a2—1'

Applying this identity with a = 32k, we obtain

1

_

1

2(32"+1)_2(32"—1)

_

1 32"+I—1'

Multiplying this by 2"”, we obtain the relation 2k+1

_

2k+1

2k+2

32k+1 ‘32k—1 ‘32“1—1' Thus 2

22

23

2n+1

n

2k+1

2k+2

m+m+m+“'+w—+1=H(W‘W)

= 1 _ 2—, n+2

32"+1 — 1

and we are done.

28

Chapter 1. Telescoping Sums and Products in Algebra

Example 1.11. Evaluate

4 . k=1 4k + 1 Solution. The key ingredient in solving this problem is realizing that the denominator factors rather nicely. Indeed,

41:4 + 1 = 41:4 + 41:2 + 1 — 41:2 = (21:2 + 1)2 — (2k)2 = (2k2 — 2k + 1)(2k2 + 2k + 1). On the other hand, we observe that the numerator 4k is simply the difference between 2192 + 2k: + 1 and 2192 — 2k + 1. Therefore

”

4k

Zn:(2k2+2k+l)—(2k2—2k+1)

k=1 4k4 + 1 = 11:1 (2k2 + 2k + new — 2k + 1) ”

1

1

3‘

=1(2k2—2k+1 _2k2+2k+1)'

We almost have a telescopic sum: letting ak = 2192 — 2k + 1, the only extra observation we need is that

2k2 + 2k + 1 = ak+1. Indeed, ak = 2k:(k: — 1) + 1 so

ak+1 = 2(k+1)k+1 =2k2+2k+1, as desired. Therefore we have a telescopic sum n

n

E 4k :2 1 _ 1 k=14k4+1 k=1 2k2—2k+1 2(k+1)2—2(k+1)+1 _1

_ and we are done.

1

_

2n2+2n

2n2+2n+1_2n2+2n+1’

Chapter 1. Telescoping Sums and Products in Algebra

29

Example 1.12. Let a1, a2, . . . ,an be positive real numbers such that alazn-an = 1.

Prove that a1

a2

a3

1 + a1+(1 + a1)(1 + a2) +(1 +a1)(1 + a2)(1 + a3) + +

an

> 2" — 1

(1+a1)(1+a2)-~(1+an) _

2"

Solution. The sum in the left hand-side is actually fairly easy to compute once we realize that ak

_

1 + ah — 1

(1+a1)-“(1+ak) _ (1+a1)---(1+ak)

_

1

_

(1+a1)---(1+ak_1)

1 (1+a1)---(1+ak)'

Therefore the sum is telescopic and

“ 1 1 1 + — (1+a1)(1+a2) + 1+a1 — —1 — 1+a1 Z k=1((1 +01)?k‘(l+ak) +

1

1

(1+a1)---(1+an_1)

(1+a1)-~(1+an)

1

- (1+a1)---(1+an)' Since

2" — 1

2n

1

= 1 _ 27’

the problem reduces in the end to the inequality 1 1 _>—_

2n_(1+a1)---(1+an)’

which is equivalent to

(1+a.1)---(1+an)22”.

30

Chapter 1. Telescoping Sums and Products in Algebra

This is a fairly standard consequence of the AM-GM inequality, since

1+al 22\/a_1,1+a2 Z 2M,.~,1+an Z 2\/%, thus (1+a1)-~(1+an) Z 2" [—01-“an :2”,

Example 1.13. Evaluate

21

22

24

28

41—1+42—1+44—1+48—1+”' Solution. We are asked to evaluate 0°

22”

0°

22"

T = Z —_

n=0 4

1

n=0 2

1

First we have

22"

22"

22"+1 — 1 = (22" + 1)(22" — 1) _

22" + 1 — 1

_ (22" + 1)(22" - 1) 1 1

-W‘H which implies N

22"

_N

1

222n+1_1_z ”=0

_

7),:

1

22"_1_22"+1_1

1

1

_1

_fi_27"fi-__1_

1 _W'

Thus, 00

22"

,

N

22"

,

1

Z 22n+1 _ 1 = 1&3)a 22n+1 _ 1 = N113; (1 _ 22N+1 __ 1) = 112:0

71:0

Chapter 1. Telescoping Sums and Products in Algebra

31

Example 1.14. (IMC 2015) Define a sequence (F(n))n20 by

F(O) = 0, F(l) = g, and F(n) = gF(n — 1) — F(n - 2) forn>2. I32 n=0

F(12")

a rational number?

Solution. Let us start by finding the general term of the sequence (F (n))n20. Note that the recurrence relation can be written

F(n) = 2F(n — 1) +

F(n — 1)

2

— F(n — 2),

in other words

F(n) _ 2F(n _ 1) = W 2 The sequence G (n) := F(n) — 2F(n — 1) therefore satisfies

C(72) _ G(n — 1 ) 2 and by an immediate induction (or using a telescopic product) we obtain 0(1)

3

G(n) = 211—1 = 2—n.

Thus

3 F(n) — 2F(n — 1) = 2—n

and dividing by 2" yields

F(n)

F(n— 1) _ 3

2—n _ W _ 47' Adding up these relations for n = 1, 2, . . . ,N and recognizing a telescopic sum on the left-hand side yields

F—éN’ =:4%=:(4%—4in)=1—4im 71:1

32

Chapter 1. Telescoping Sums and Products in Algebra

which yields

F(N) = 2N — 2—”. It follows that

1

_

1

_

22"

_

1

1

Fan) _ 22" _ 2—2" _ 22n+1 _ 1 _ 22" _ 1

22n+1 _ 1'

We recognize the telescopic sum from the previous example and we conclude that

Z— F(12")= All—13100” 1112— 0F(12n)

n=0

"ON

.

1

1

= 13.310020 (221; _ 1 — 22n+1 _ 1) n:

1

.

=$a0*afifi:>=L which is a rational number.

Example 1.15. Let

h—1+1+ 1+ n_ 3 5 Prave th a t:

1

+ 2n— 1 forn 6 N*.

1

1

fi+w+m*fififi 1

) _

Then, using

.

cota +cotfi = W, sm 0: sm ,8 the sum becomes , cot

27r

37r

27r 2 2 7+cot 7 +cot 7 =3+2

21r __

_ cot7cot 7

37r _

cot 7

37r

sm— —7 7r,

27r

Sln—Sln—

7

coszcosz—7r — cos3—7r __7____7 __ -3+2 7r 21r SlIl—Sll'l—

7

7

7

Chapter 2. Telescoping Sums and Products in Trigonometry

cosgcos? —cos (; + 2%)

69

sinfsinifl

=3+2———7

7

=3+2

sinzsin2—7r sin—sin2—7FZ5' 7 7 7 7 Note that this equality can be seen as an instance of the general formula

'n,

Zootz 16:1

(2n + 1) = 3 = n( 2 n _ 1)

km 2n + 1

(2n + 1)

3

1

(for n = 3 we get our result). This general formula holds because as we will see in Example 3.5 in the next chapter

are the roots of the equation n

g(

—1 7"

2n+1

) (2r+1)x

"‘7‘:

0

(and we can therefore evaluate their sum by Vieta’s formulae). Thus, in our particular case, 7r

27r

37r

cot2 7, cot2 7, cot2 7

are the roots of the equation 3

7 E :(—1)"(2r + 1):!33—T = 7x3 — 351:2 + 21:1: — 1 = 0 r=0

which proves once again that their sum is 5. Example 2.20. Show that 7r

27r

37r

tan7tan7tan7 — W.

70

Chapter 2. Telescoping Sums and Products in Trigonometry

Solution. We have

tan7t_ 7tant—35tan3t+21tan5t—tan7t _

1 —21tan2t+35tan4t—7tan6t

(why? again, see the next chapter for a clever way of proving this), which

yields (by replacing t with 7r/7, 27r/7, 37r/7, respectively, and by using the fact that the tangents of these numbers are not zero)

7—35x+2la:2—x3=0®x3—21x2+35$—7=0 2 3 for at: E {tan27r 7, tan2 7“, tan2 7’”

As the numbers from this set are mutually distinct, they must be all the solu-

tions of the above equation, therefore their product is 7 (by Vieta’s relations): 7r

27r

37r

t 8.11 2—t 7 an 2—t 7 an 2— 7 =7. Since all the tangents involved are positive, the conclusion follows. Example 2.21. Show that , 27r 7r 4sm7—tan7 —-\/'7.

Solution. Again, by

tan7t_ 7tant—35tan3t+21tan5t—tan7t _

1 —21tan2t+35tan4t—7tan6t

we find that the six roots of the equation

x6—21x4+35x2—7=0 are :1: tan 7r/7, :1: tan 27r/7, :Iztan 37r/7. Now, since

x6—21x4+35x2—7=(x3—7x)2—7(x2+1)2 =(x3 +\/_:v2 —7:1:+x/_)(:r:3 —\/—x2 —7x—fi),

Chapter 2. Telescoping Sums and Products in Trigonometry

71

we have that 'u—tan7r

‘

7

is a root either of

P(a:) =ac3 + \/7:z:2 —7m+\/7, or of

Q(:L') = x3 — V7332 — 7:1: — \/7 Because

P(0)P(1) = W (2x/7 — 6) < 0

P has a root in the interval (0, 1), and that can only be tan 7r/7 (from the numbers :Iztan 7r/7, :|:tan 277/7, :lztan 377/7 this is the only one between 0 and 1). Thus we have u3+\/7 2—7u+\/7=0 which can be read as

8u __ _

1 +u2

=

u

7

f’

meaning that 2

4 sin _7r — tan I = \/7 7

if we use

7

. 2tana sm2a = —-——2—— 1 + tan (1

Example 2.22. Show that

f/cos—+\/;)S—+\/coss71r =\3/% (5— 3W). Solution. This 1s a famous identity of Ramanujan. We have cos2—7r +cos4—7r+c038—7r —cos2—7r —cos§£ —cosE — ——

7

7

7 _

7

7

7 _

as we showed in Example 2.20. Then 87r

cos 27‘- cos 47f + cos 2” cos 8—” + cos 4—” cos 7 7 7 7 7 7

2

72

Chapter 2. Telescoping Sums and Products in Trigonometry

—1

cosG—W+cos2—W+cos107r+cos6—7r+cosl2—7r+cosfl

'2

7

—1 _2

7

7

7

—cos:+cos2—7r—cos31r 7 7 7 ——cos:+cos27r

._

7

7

7

7

cos7r+cos2—7r—cos3—7T 7 7 7 cos37r—

and

7 _

,

1

2

161r

cos 27f cos 4” cos 8” — sm 7 7 7 —8'

7 — 1 27f _8'

$1117

Thus, letting I l 4 I a: 3 2003273, b= 3 2cos77r, andc= 3 20058771we have for (13, b3, and c3 the equations a3 + b3 + c3 = —1, a3b3 + be‘c3 + 030,3 = —2, (13b3c3 = 1 => abc = 1. Let cv=a+b+cand y=ab+bc+ca. Then

a3+b3+c3—3abc= (a+b+c)(a2+b2+c2—ab—bc—ca)

=> 73(932 — 3y) = —4. In a similar way (with ab, bc, ca instead), —5 = (ab + be + ca)((ab + be + ca)2 — 3abc(a + b + 0)) = y(y2 — 3:13). To solve the system for m, use the first equation to get

_:1:3+4 y—

3:):

and substitute. We get

(9:3 + 4)((a:3 + 4)2 — 27x3) _5 =

271:3

Chapter 2. Telescoping Sums and Products in Trigonometry

73

0 = 135.133 + (x3 + 4)(:1:6 — 199,;3 + 16) = x9 — 153:6 + 751133 + 64 = (x3 — 5)3 + 189

=>(5—x3)3=7-33=>:c= 3 5—337? as desired. Example 2.23. Prove that

sinl+sin3l+sinfl=l

13

13

13

13+3

2

2

13

'

Solution. After squaring both sides (which are positive) and after transforming products into sums we need to prove that 27r 671 SW cos — + cos — + cos— 13 1313

_2

OS4_+COS 1071' HOS 1211' 13

13

—1+3‘/fi.

13

4

Let cos2— +cos67r +cos8—7r— :1:

13

13

13 _

and cos4—7r+cos1—O—+cos 127T —

13

13

1—3 _ 1"

Then we compute that

1 3 x+y=~§ andxy=——. 4

Indeed, the first of these is a. special case of the next problem, since we have _ 6

67f cos T3 77f 2k1r _ sin —3

x+y—;cos 13 —

Sinl

—

_

13

s1n 6—71. cos 6”

13

sin —

13=_l. 2

s1n 127T

13_ 1.

sin 1

2

74

Chapter 2. Telescoping Sums and Products in Trigonometry

The second follows by expanding and repeatedly using the formula

2 cos acos b = cos(a + b) + cos(a — b). We get 2x

47f +cos 10” y ‘— eos6—7r 13 +cos 27‘13 +c°s 12% 13 +cos 87F 13 +cos 13 13 +cos10—7r +cos2—7r +cos16—7r +cos—7r +coslS—7r +cos 67f 13 13 13 13 13 13 +cos12—7r+cos4t—7r+coslg—7r+cos2—7r+cos207r +cos47r 13 13 13 13 13 13

— 3

_

cos2—7r +cosfl +0039: +cos8—1r +coslo—7r +cos12—7r

13

13

13

13

13

13

(where we also used cos(27r — a) = cos a), hence

a: y — + y) — — §(:c 2 — —§ 4Thus (as a; is positive and y is negative),

x/fi — 1

«13 + 1

4

4

a; = — and y = —-—..

Consequently, 00527: +cos6—7r +cos 87f 13 13 13

2

m4

=

4

cos4—7r +coslo—7r +cos12—7r 13 13 13

«EH

+

2

1+3x/13

=

4

'

Example 2.24. Evaluate 'n,

2 cos kw. k=1

Solution. Assuming that a: aé 2m7r, m an integer, we multiply by 2sin (17/2. From the product-to—sum formula we get TL

n

2gsingcoskx=z (sin (k+%)x—sin (ls—é) as) k=1 1 1 =sin (71+ 5) x—sin; = 23inn7flcosw.

Chapter 2. Telescoping Sums and Products in Trigonometry

75

Therefore we get 1 n

2—x2sm§

Zeoskx= _.—?II—_ —§ = k=1 2sm—

2

Clearly, when a: = 2m7r, m is an integer, the answer is n.

Example 2.25. Prove that, for all real numbers :10 7E 2m7r (with integer m), we have sin

E s' (n+ 1)ac 1n

sinx+sin2x+sin3x+---+sinn:1:=

a: sin — 2

2

Solution. As in the previous problem, we multiply the sum with 23m x/2,

and use formulae for transforming products into sums and vice-versa: . x

n

'n

.

n

. a: .

1

1

2s1n§ Zsmkx = Z2sm§smkw = 2 (cos (k — 5) a: — cos (k + 2) 33> k=1 k=1 k=1

_

x

— cos 2

cos

(2n+1):1:_ 2

. n56. (n+1)$

— 23m 2 sm

2

,

and this is what we had to prove. Of course, for a: = 2m7r the sum is 0. Note that mathematical induction can be used in such problems, too. Also, observe (and prove) that the slightly more general identity .

na ,

sm—sm

sinx+sin(a:+a)+~~+sin(:1:+(n—1)a)=


0 be two

sequences of positive numbers such that ylyg-nyk lxg-uxk for all 1 S k Sn.

Prove that

y1+y2+---+yn Z$1+$2+-'-+xnSolution. Combining the hypothesis and the AM—GM inequality, we obtain fl+y_:+n +y—k>k’= y1y2”'yk $1

$19

$1$2“‘$k

2k.

On the other hand, Abel’s summation formula yields (we let xn+1 = O)

k

y1+y2+'+yn=ZZ—kk $k=k:($k—$k+1). k:1 =1 By assumption xk—k Z 0 for 1 g k S n, which combined with the previous observation yields

y1+y2+

+212

91:

+yn=k;(mk—$k+l)(:—: +x—2+ ' +217) n

ZZk(mk—xk+1)=x1+m2+-~+xn k=1

Chapter 4. The Abel Summation Formula

111

and finishes the proof (for the last equality Abel’s summation formula was

used once more). Example 4.6. Let a1, a2, a3, . . . ,an be positive real numbers such that

1 (1102 ' "(1k 2 W for all 1 S k S n. Prove that 1

1

1

a1+a2+---+an2 n—H+m+- --+%. Solution. We choose yk = ak and wk:

__L____i__l_ (212— 1)2k ‘ 2k — 1

2k

in the previous exercise. We clearly have :01 2 1:2 2 -- - 2 55,, and, by hypothesis, —aa Zlk— 12

19192

ak_(2 > 1k)l-1 — '2 1

1 3.4

(2k—1 1)2k

=$1$2...xk

for every 1 S k S n.

So the result from the previous example applies, and we deduce

y1+y2+"'+ynZ$1+5I32+"'+$m thatis,

a+a+

1

+a >1—1+1

2

”-1

—1+1+1+1+ ‘1

2

3

4

+

2

2n—1

+2

_1+1+1+1+” + ‘1

as required.

2

3

4

1+-- +

4

2

1

n+1

n+2

1

1

2—72 +1

_1+1+

+1

1

+

1

1+1+

2

2n

__1__+

2—n

1 n+n’

4

2

2n

+71,

112

Chapter 4. The Abel Summation Formula

Example 4.7. Let a1,...,an be real numbers such that a1+---+akSk for lnn. Prove that

1 an __ 1. By the assumption we made we have an+1 = n! +

n. 2

n.

and our claim is proved: we have an = n! for every natural number n.

Example 5.8. Let m1 = —2, x2 = —1 and

xn+1 = \3/n(:z:.,2, + 1) + 2:1:n_1 forn Z 2. Find $1 +1132 +

+122009.

Solution. One sees that $3 = 0 and x4 = 1, thus one may guess that an = n—3 holds for all positive integers n. We prove this by inducting on n; assuming that the result wk = k — 3 is true for all k g n (in particular 33,, = n — 3 and xn_1 = n — 4), we will have, by the given recurrence formula,

xn+1 = \3/n((n — 3)2 + 1) + 2(n — 4) = \3/n3 — 6n2 + 12n — 8 =n—2=(n+1)—3,

Chapter 5. Mathematical Induction

129

completing the proof. In this situation :21 + x2 + - - - + x2009 = 2009 - 1002 = 2013018 follows quickly, doesn’t it? Example 5.9. Find the positive real numbers 9:1, :32, . .. that satisfy

fi+d+m+fi=m+m+m+nf for any positive integer n.

Solution. We have at? = is? (for n = 1) and x1 > 0 yielding 51:1 = 1. Then, for n = 2, we get

1+x3 = (1+x2)2®w§—x§—2m2=0©$2(x2+1)(x2—2) =0, hence (being positive) :62 = 2. These results, and the well-known identity 13+23+~~+n3=(1+2+---+n)2 are good reasons to believe that wk = k for all positive integers k. Supposing

that this is true for every k 6 {1,2, . . . , n} we will have (by hypothesis)

fi+21+~swfi+x54=(r+2+~-+n+nH02 nn+1

2

nn+1

2

(l) xii+1 — 333;“ — "(n + 1)90n+1 = 0

(it $n+1(93n+1 + n) (93n+1 - (n + 1)) = 0, therefore xn+1 = n + 1 follows, finishing the inductive proof. Example 5.10. Let a1,a2, . . . ,an be distinct positive integers. Prove that

fi+fi+m+fi2

2n

;(m+t+m+%i 1

Solution. The statement is clearly true for n = 1 (it becomes a? 2 of), so let us assume it holds for any n distinct positive integers, and prove it for n + 1. Thus, we want to show that

fi+fi+m+fi+fiflz

2n+3

3

(a1+a2+~-+an+an+1)

130

Chapter 5. Mathematical Induction

is true for any n + 1 distinct positive integers a1,a2, . . . ,an+1. Due to the symmetry we can assume that al < a2 < - - - < an < an“.

We have, by the induction hypothesis, that

a§+a§+m+ai2

2n+1

3

(a1+a2+-~+an).

If we could prove that

2 2n+3 a3“ 2 5(01 +a2 + ' ' ' + an) + Tan+la adding these inequalities would yield the desired conclusion. Thus, it remains to show that 2

2n + 3

2

an+12 3(01+02+-"+an)+

“n+1:

3

for any positive integers a1 < a2 < - -- < an < an“.

Because they are integers, the above inequalities actually say that “n+1 2 an + 1; an+1 2 (111—1 + 2,

and so on, until an+1 2 a1 + n. Consequently, 2 2n+3 an+1 3(a1+a2+---+an)+ 2

S§(an+1—n+an+1—(n—1)+"'+an+1-1)+ 2

+1

3

2

2

2n+3

3 3

“n+1 -

And now we have

2

n(n+1)

5 (nan+1 — T) +

2n+3

3

2 an+1 S “n+1

because it is equivalent to

2 o, (am — (n + 1)) (“n+1 — fl) 3

3

an+1

Chapter 5. Mathematical Induction

131

which is true, since an+1 2 a1 + n 2 n + 1. Our proof is done. Try to prove that

nn+1 a%+a§+...+a32¥ 2 (a1+a2+-~+an) for any distinct positive integers a1, a2, . . . ,an. Example 5.11. Prove that the inequality at” — n11: + n — 1 2 0 holds for any positive real number a: and any positive integer n 2 1. Solution. We show that

(1+a1)(1+a2)---(1+an) 2 1+a1+a2+~~+an for any real numbers a1, a2, . . . ,an all greater than —1 and all having the same

sign (possibly, some of them are 0). The base case n = 1 is obvious. If we have

(1+a1)(1+a2)~-(1+an) 2 1+a1+a2+~~+an we can multiply in both sides by 1 + an+1 and get

(1+a1)(1+a2)---(1+an)(1+an+1)Z(1+a1 +a2 +~'+an)(1 +an+1) =1+a1+a2+---+an+an+1+an+1(a1+a2+~-+an). Since all numbers a1, a2, . . . , an+1 have the same sign, we have

an+1(a1 +a2+---+an) 20 and the conclusion follows for n + 1 numbers. Now let a: > 0 and consider a1 = a2 =

= an = a: — 1 which satisfy the

conditions for the above inequality. And the inequality reads

(1+x—1)"21+n(a:—1)0 (+n+1> n (+n+1)+ n (i)

1

l+—

n+1

Eli

n

1

1

>1+—¢)

1+——

n

n+1

”+1

>

1 n

1+—

n

.

That is, Bernoulli’s inequality implies the monotonicity of the sequence with general term 1

n

(1 + —) n

,

which defines the number e (as being its limit). Example 5.12. Prove that n—l

An=Zrkcoskt= k=0

l—rcost—rncos t ”+1005 —1 t 11. +7" 2 (n ) _ 1 27" cos t + r

and 71—1

3,, = 27"“ sinkt =

k=0

r sint — 7‘" sin nt + rn‘l'l sin(n — 1)t 1 — 2r cost + r2

Chapter 5. Mathematical Induction

133

for any real numbers r and t with 1 — 2r cost+r2 aé O and any positive integer n. Solution. We already have shown that these equalities hold by using de Moivre’s formula. Now we prove by induction the formula for An and leave Bn to the careful reader. For n = 1 we have to see that A

:1:

1

1 —rcost—rcost+r2 1—2rcost+r2

’

which is clearly true. Then we have (supposing that the formula for An holds) A

”+1

=A

n t= "+7. COS”

1 — rcost — r" cos nt + r"+1 cos(n — 1)t 1—2rcost+r2

+ r” cos nt

_ 1—rcost—r"cosnt+r"+1 cos(n — 1)t + r" cos nt — 2r""’1 cos tcos nt + r"+2 cos nt 1 —2rcost+r2

_ 1 — 7' cost — r”+1 cos(n + 1)t + r"+2 cos nt 1 _ 1 — 27‘ cost + r2 because

r7“?1 cos(n — 1)t — 27"”"'1 cos t cos nt

= rn+1 (cos t cos nt + sin t sin nt — 2 cos t cos nt) = —r"+1 cos(n + 1)t. Thus the formula for An+1 holds and we are done. Example 5.13. For each positive integer n and each real number :1: prove the following inequality

n I cosa: I + I 00529: I + I cos4a: I +---+ I c032nx I 2 —. M Solution. First we observe that the inequality

> i Ial + |2a2 — 1| _ fl holds for any real number a.

134

Chapter 5. Mathematical Induction

Indeed, this clearly holds for |a| 2 1/\/§, while for |a| S 1/\/§ it can be written in the form 1

|a|+1—2a22 —.

\/§

For example, when a E [0,1/x/2—J we have to prove that 1 f(a)=1+a—2a22——.

fl

But f is a quadratic function with maximum attained at a = 1/4 E [0, 1/\/fl (the maximum is 9/8, but this is not important in this matter). Therefore, the minimum value of the function in the interval [0, l/x/fl is

{(7)}? Thus

|a| + l2a2— 1| =f(a) 2 % follows for a E [0, 1 / m , too. The reader will surely be able to prove the inequality in the case a E [—1/\/§, 0] by a similar reasoning (or just by noting

the parity of the function a I—-> |a| + |2a2 — 1|). In particular, we have

| cost l + I cosZt |

1 = | cost | + | 2cos2t—1 | 2 — fl

for every real t, implying that P(1) is true (it is weaker than the above inequality), if we denote by P(n) the statement of the problem.

For P(2) we have 1 |cos {13 l+|cos2m|+|cos4x|>|cos x |+|cos2ac|> _

_

_ x/i

2

= _,

2\/§

according to the same inequality proved above.

And now we show that P(n — 2) implies P(n). Indeed n

1

n— 2

Z|cos2kx| = Z|cos2kct| + |cosZ” 1ar|+|0032nsc| >n_2+

k=o

16:0

n

= —-

2:5 5 W5

Chapter 5. Mathematical Induction

135

and the inductive proof is done. Note that the following proof is also available

(also based on |cost| + |cos 2t| Z 1/ fl) Adding the inequalities 1

|cos2k‘1x|+|cos2km| 2 — fl, k = 1,2,...,n

(together with |cosm| Z |cosx| and |cos 27%| 2 |cos 2”ac|) yields 11

2Z|0032kx| Z l + |cosm| + |cos2nz| fl

k=0

and since the sum of absolute values from the right side is nonnegative, we obtain the desired inequality 71,

2

n

|cosk| Z —.

It is a matter of taste if we choose the proof that uses induction, or the one that avoids it. The most interesting fact about the two proofs above is that there is one simpler than each of them. We are sure that, carefully analyzing the cases n = 1 and n = 2, the reader will be able to find it. The lesson to be learned here is that a proposition depending on a positive integer variable need not be proven by induction. Example 5.14. Prove that Jackson’s inequality: , sin 2:1: smx +

---

sin nx

>0

holds for any a: E (0, 7r). Solution. Let

,

Sn(:l:)=s1n:1;+

. sm 2:1:

2

+---+

For n = 1 we clearly have

51(18) = sinac > 0

_ s1n nzz:

n

.

136

Chapter 5. Mathematical Induction

for all a: E (0,7r). Assume that Sn(x) > 0 for every x E (0, 7r) and let us prove that the same is true for Sn+1(a:). The derivative of Sn+1 as a function of a: is sm sin —

2

and has zeros (hence the critical points of Sn“)

x’ _ 2k7r

k_n+1 and

x” _ (2k + 1)7r k_

n+2

for 0 < 2k < n + 1 (since we want as;c and 23;; to be in the interval (0, 7r)). The extremum points of Sn+1 are among these points and the endpoints, and we have, by using the inductive hypothesis,

sin(n + 1):z:;c Sn+1 (1:2) = 512062) +

_

I

and sin (n + 1)(2k + 1)7r

sin(n + 1)a:”

SWIM = SW"; + TU” = SW35) + +2 II

.

(2k + 1)7r

Sln

n+2

//

2

Sin ml];

=Sn(.’13k)+—n+-i——— = Sn($k)+ n+1 > 0

because 33’]; E (0,7r). Also, the limits of Sn+1 at the endpoints 0 and 7r of the interval are both equal to 0, therefore the conclusion follows. We invite the reader to prove that n

Z(_1)k+1sinkx > 0 k=1

k

for every x 6 (0,71') (not necessarily by induction, but rather as a consequence

of the result that we have proved).

Chapter 5. Mathematical Induction

137

Example 5.15. Prove that

2x322":k=1

Solution. Let Sn and Tn denote the left-hand sum and the right-hand side respectively. We have

fink“ ) (19)) -—n+1+;z(k—1> n+1

Sn+1—

1

n

:1(n+1)=

1

1

n 1

2n—2

n

271—1

n+1+n+1=n+1'

=n+1+n—+lk

In the previous calculation, we used the recurrence formula of the binomial coefficients, then the formula 1

n

k

k—l

_

n!

_

1

—k!(n+1—k)!_n+1

n+1 k

’

and finally the sum of the binomial coeflicients from the (n + 1)—st row of Pascal’s triangle. On the other hand, it is clear that Tn+1 —

Tn

2” — 1 = —. n+ 1

Now we have 81 = T1, and if Sn = Tn we also get S=n+1

(Sn+1— Sn) + S

= (Tn+1 — Tn) ‘l' Tn = Tn+1

and the identity follows for all positive integers n.

Example 5.16. (USA TST, 2000) Let n be a positive integer. Prove that

n0 _1+ "1 4+

+ nn _1—n—+13+g+ ‘2n+1 1 2

+2”+1 n+1'

138

Chapter 5. Mathematical Induction

Solution. The proof is by induction on n. Denote the left hand side by an, and the right hand side by bn. The base case is clear and we only need to show that an and bn satisfy the same recurrence. A recurrence for bn is simple: n+ 1 2n

bn = —— n—l + 1-

So we only need to prove that an =

We have

n+ 1 an_1 2n

+ 1.

1

n+1 n—1 _ _(n+1)i!(n—i—1)! 2n

2'

_

2(n!)

To express the right hand side in terms of binomial coefficients of base n, we

write n + 1 as (i + 1) + (n — i) and conclude that

n+1 n—1 ‘1

W( i )

((i+1)+(n—i))t!(n—i—1)!

=

2(n!)

=%((i:1)—1+(7:>—1)' By summing these relations and using the fact that

(3)=(:)=1, we get

and we are done.

Chapter 6

Combinatorial Identities and

Generating Functions Combinatorial identities deal with binomial coefficients. The binomial coeffi—

cients (Z) are defined by n n! n = 0, otherw1se, < < 'f — = (k) and n, _ k — 0 1 k!(n _ k)! (k) where n is a nonnegative integer, while k is an integer. Remember that 0! is

defined to be 1. A very useful relation is the recurrence formula of the binomial coefficients: n+1

_

n

k+1 _ k

+

n

k+1’

which you can check applying the definition. The binomial coefficients can be

found using the binomial theorem (Newton’s formula) (a + b)” = Zn: (2) an_kbk. k=0

Many useful formulas are easy consequences of the binomial theorem.

140

Chapter 6. Combinatorial Identities and Generating Functions

For example, the above formula with a = b = 1 yields

2(Z)=(3>++-~+x+~-+(nil)x”_1+(:)m"+(n:1)z"+1+--- ,

because (2) = 0 for k > n. The function (1 + an)” is called the generating function of the sequence ck = (2). For every sequence (0101120 we can associate

an infinite series

F(a:)=a0+a1x+a2x2+~~+anmn+~~

called the (ordinary) generating function of the sequence. As an example let us take the sequence az- = 1, then

F(a:)=1+m+a:2+---+x"+---= Thus

1

l—x'

Chapter 6. Combinatorial Identities and Generating Functions

141

is the generating function of the sequence ai = 1. Let us list some of the most important sequences and functions corresponding to them: The sequence

ak = 1 /k: corresponds to 1

1

—ln(1—x)=a:+§x2+§x3+~-, from this result it immediately follows that 00

_

1112—2:

(—1)”-1_

Z.

_1

1

1

2+3

1

4+

n=1

The series corresponding to the sequence ak = 1/k! describes the universal constant e:

$3 + ez _— 1 + ha: + ”02!—2 + i For the Fibonnaci sequence (F0 2 F1 = 1 and Fn =

n_1 + Fn_2 for n 2 2)

the generating function is 1 m=1+$+2$2+3$3+5$4+m

and, of course, Newton’s formula says that the sequence ak = (2‘) corresponds to: 00 (1 + x)“ = Z (3)33”. n=0

Example 6. 1. Evaluate

142

Chapter 6. Combinatorial Identities and Generating Functions

and

(2)42) + (z)

2 (2:11)

we have 11.

A+B=Z(7> =(1+1)"=2'n,

i=0 3 A—B =j§:%(—1)i(;‘> = (1— 1)“ = 0, therefore A = B = 2”_1. These are also fundamental identities satisfied by the binomial coefficients. Example 6.2. Prove that

Enjkcc‘) =n-2”_1. Solution 1. Because

I“ (Z) = kwnni k)! = “(k fin—(7:): k)! = ”(2: i) we get

2(2)=in=n§(“;l)=n-2n—a k=1

according to the fundamental identity of the binomial coefficients. Solution 2. We start with the binomial formula in the form

:0 (DJ = (1 + as)", valid for any real number x. Differentiating with respect to ac yields n

E k (2) ask-1 = n(1 + x)”_1, k—l

Chapter 6. Combinatorial Identities and Generating Functions

143

which is a more general identity — it is enough to take at = 1 in order to obtain the result stated by the problem. Try to prove in a similar manner the identity

I: k(k — 1) (Z) = n(n — 1) . 2"-2 (use k(k — 1) (Z) = n(n — 1) (Z : 3) or differentiate twice), then infer n

2 k2 (2:) = n(n + 1) - 271—2, 16:1

for any positive integer n. Also prove that for n 2 3

i(—1)’%(Z) = En:(—1)kk2(’l:) = 0. Example 6.3. For nonnegative integers m and n show that

§'°(Z>=m(";1)Solution. The result is clear for m = 0. Assuming that it holds for some m, we will have m+1

[cg—1r (Z) = genkc) + (—1)m+1(m: 1) = (—1)m(n;l 1) + (—1)m+1(m:‘r 1)

= (-1)“ H”; 1) + (77511)) =

_1m+1

(

)

n

1

(m+1 ’

144

Chapter 6. Combinatorial Identities and Generating Functions

as we needed to prove, according to the recurrence of the binomial coefficients. Note that the result from the right—hand side is 0 for m 2 n 2 1. Actually, for m 2 n 2 1 the identity is always the same, namely

gang) = 0, which by the binomial theorem basically says (1 — 1)” = 0. Also observe that a direct proof (by telescoping) can be obtained if we replace each (2) with (n?) + (:3), by the same recurrence relation for the binomial coefficients.

Example 6.4. For positive integers m and n with m S n prove that 11,

Z k=m

k

n

m

k

= 2n—m

n

.

m

Solution. We have

mom—Wfi n!

(n—m)!

m!(n — m)! ' (n — k)!(k — m)!

= (2X11?) i (51) (Z) = i (2X21?) = (2:) Z (”3’") = W371)-

k=m

k=m

j =0

We changed the index of summation with the substitution j = n — k, when 19 runs from m to n we see that j runs from n — m to 0. Example 6.5. Evaluate n

1

Sn=:2—k(

k=0

n+k k

).

Chapter 6. Combinatorial Identities and Generating Functions

145

Solution. We have So = 1, 51 = 2, 82 = 4 and 8'3 = 8, therefore we have reasons to believe that 5,, = 2“ for every: nonnegative integer n. We can prove this by finding a recurrence relation for Sn. Note that (

n+1+k ) k

and

n+k n+k > (k )+(k_1)fork_1

=

1

0

0

(n+0+ )=(n3— >fork=0.

Thus

“

+1

1

n+k

”

+1

1

n+k

Sn+1=z27( k )+Z2—k(k—l> k=0 k=1 ”i1 n+k k=0

Since

2k

1

2n+1 #51 n+k _

+2n+1

n+1

k=1

2k k—l

1

2n+2

271+2

n+1

l 2n+2

_l(2n+2)(2n+1)!_

2n+1

2

_2(n+1)n!(n+1)!_

n+1

n+1

‘

the two terms extracted from the sums above cancel, and we get n

1

n+k

n+2

1

n+k

Sn+1=2fl( k )+Z2_’°(k—1> k=0 16:1

_8 +1n+2 1 _ "

2H2k-1

(n+1)+(k—1) k—l

1 =Sn+§Sn+la

that is, Sn+1 = 2.3,. Now an easy induction (based on this recurrence relation) shows that, indeed, 5,, = 2” for all n 2 0. Example 6.6. Prove that

(3)3 (3211+ (3)2: (2:)

146

Chapter 6. Combinatorial Identities and Generating Functions

Solution. Consider the function f (so) = (1 + 1102“. From the Binomial Theorem the coefficient of :3" in f (x) is equal to (2:). On the other hand, we have

(1 + m2” = (1 + 11:)"(1 + x)“

= ((3)+(T)m+-"(Z)w"> ((3)+(’I)x+-~(Z>x")If we perform standard multiplication, the coeflicient of x" will be equal to

enragwwaaR%®%£mw (3)24—(:)24-n.4.(:)2==(if). Example 6.7. Evaluate

n0 + n4 + n8 + Solution. Let us denote by A the sum from the problem statement, and let us consider also n

n

n

B‘(J+(9+(J+”” n

n

n

C“ (2) + (6) + (10) +"" and

wownmww

Chapter 6. Combinatorial Identities and Generating Functions

147

Then for n _>_ 1 we clearly have (by the binomial expansion) 17.

A+B+C+D=Zk

(n— k)

(n+k)

_ 1

n 2” (—W‘

2W}; T+T ——n+(—1)Z _

-

'

Now the formula

1

2n+1

(2n) = 2n + 2

j

1

+

1

(2n + 1)

(2n + 1)

j

j+1

can be easily checked by using standard computations (and we invite the reader

to do that). So

222:5

mm j

j+1

150

Chapter 6'. Combinatorial Identities and Generating Functions

_ 2n + 1 2"

(—1)J'

_ 2n + 2 H

(2n + 1)

_ (—1)J'+1 (2n + 1)

j

j+1

_ 2n+1

(—1)0

_ (—1)2n+1

_2n+2

2n+1

2n+1

0

2n+1

_2n+1

_n+1’ leadingto Sn

1

2—:——

(2n)!

2n+1

—1"-—.

2n +( )

n+1

n Finally we get

1 (2n+1)! .S'n =_2(n.)I2 +( _ 1) n.— 2(n+1)' It hardly seemed in the beginning that we would get a telescope for this sum, did it? Solution 2. This is more involved, but showcases a useful trick. As in the first solution, we have

Sn _ 2n: (-1)’° (2n)!

k=0 ( 2n > n—k

Now, the reciprocal of a binomial coefficient can be expressed by an integral with the following formula: 1 1 . . _ (m) = (m+1)/0 a:J (1 _ ac) m-J dx,

i

Chapter 6. Combinatorial Identities and Generating Functions

151

and consequently we have

(271n)‘

=(2n+1)Z(—1)k/ w"_k(1—x)”+kd:z:

=(2n+1)/01:1:x"(1—$)"kz=0(m ;I)kd:c, where we arrived at the sum of a geometric progression:

i(x_1)k= (967—1)n+1—1= $n+1_($_1)n+1 k=0

a:

”‘T‘l—l

x"

_ xn+1 + (_1)n(1 _ m)n+1

_ f: We get further

(21:)! (2n+1) / (:1:"+1(1—:1:)"+(—1)”(1MM“) _2n+1

1

_2n+2

(2n+1)+( 1)

n

’

n

1 where, for / xn+1(1 — as)”dcc we also used the formula from the beginning, 0 with m = 2n + 1 and j = n + 1 (and we wrote (2";1) instead of (2133)) Finally, after multiplying by (2n)! and a few more calculations, we obtain our result

|

n. 8,, = 3(n! + (—1)"(n + 2) - . - (2n + 1)).

To prove the integral formula for the reciprocal of a binomial coefficient, notice

that we have (integrating by parts)

1 Ij =/ xj(1 — 51:)”‘jdx o

152

Chapter 6. Combinatorial Identities and Generating Functions

=—mj+1(1—x)”_j‘:+j—j+1/O xj+1(1— x‘)” , 1am, 3+1 thus

_+1 .7

Ij+1 = nTj-Ij.

The desired formula follows by iterating this relation, until we arrive at 1

I =—

0

n + 1’

which is easy to check. Example 6.10. Prove that

3

.5

it: ‘1)=%-

Solution. We have

”‘l't’lotib —f—1(:: i> ‘%(;::i>‘ 2(z+1)+1 i=1 2t 2(z+1)

_1+1_3 ‘1

2—2

We can also evaluate the final sum directly as a telescope with step 2: 00

[9:

__;L___ nl—mom§:_______ hm _i___L_ 2(ls—1)(k:+1) 2(look=2 k—l k+1 n

—lim 12+1 _n—>oo 2

1 n

1 n+1

—1+1—3 _ 2—2'

Example 6.11. Evaluate 1 m+n

22(— min! ' Solution. Observe the following fact

22(g—"ZZF—(z_“—m13m)(z x+fi+m+fl3

+(—_—_)_...+(_1

3

w4a+fi+m+flw. n

156

Chapter 6. Combinatorial Identities and Generating Functions

By the multinomial theorem we have (11: + 1122 + _ . _ + xn)p =

2 :

(.71 +j2 + - . . + j")!$j1+2j2+-~+njn

‘ l' I... ' l

j1+j2+---+jn=p

‘71‘72'

'7'"

hence the coefficient of as" in this formal series will be

(_1)j1+j2+-~-+jn—l (jl +j2 + - - - + jn)! ‘l 'l‘l...311,]1 Jl-JZj1+2j2+m+njn=n 2

—

Z

_ j1+2j2+~~+njn=n

1 +32 +

-

+ 37;,

1M_ (-1)jl+j2+---+jn— ' I' l... .7n' l .71-32-

Since first we found this coefficient to be 1/n, the desired equality follows and our proof is done.

We close this chapter with the classical (and very beautiful) theorem of Lucas which evaluates the binomial coefficients modulo a prime. Example 6.14. Let p be a prime, let n and k: be positive integers (n 2 k), and letn=n0+n1p+~-+n3ps andk= k0+k1p+'--+ksps be the basep representations of n and k. Here n,- and k, are base p digits, that is, they are

all from the set {0, 1,. . . ,p — 1}. We then have

(3:) (3:3) (2:) (2:) W"

Solution. In the proof we will work in the ring Zp of integers modulo p (and, more precisely, we will work in the ring Zp[X] of polynomials with coefiicients in Zp). The elements of Zp are equivalence classes of integers, but we will not use specific notations for the equivalence classes. Instead equivalence classes will just be denoted by numbers. Since we are working in Zp, we will have equalities rather than with congruences. Thus, for example, the congruence

(93 + 3/)” E 56" + 3/” (mod 10) (proved in the chapter Mathematical Induction) will be written

(x+y)p=w”+y”,

Chapter 6. Combinatorial Identities and Generating Functions

157

because now a: and y actually denote the classes of equivalence of x and y modulo p. In fact this is the starting point of the proof. It is clear that

(f+g)”=fp+9p holds for f, g e Zp[X], too (the proof is the same); moreover, we have

(f + g)“ = fp’ + f1 for any positive integer j (we leave the proof by an easy induction to the

careful reader). Thus, in Zp [X] we have

(1 + X)” = (1 + X)"°((1 + X)P)”1 - - - ((1 + XV)?” = (1 + X)”°(1 + XP)"1 - - - (1 + XP‘)”s no

= 2 (7°) X10 - _

30—0

.70

n1

n,

2 (70x11? ... Z (7) W, . _

11—0

.71

- _

13—0

.73

Now the coefficient of Xk in (1 + X)n is, of course, (2). The expansion of the final product from above contains all terms of the form

(no) ("1)

(”S)o+j1p+-~+jsps, .78

.71

.70

with 0 3 ji 3 ni for 0 S i _ 1, then p2 | mn and so n(m) = a(mn) = 0 and we are done. Similarly if p2 | n we are done.

166

Chapter 7. Sums and Products in Number Theory

Otherwise, m = p1 ' - -pk, n = q1---qh, where p1,...pk,q1,...qh are distinct

Primes- Then #(m) = (—1)k, M”) = (—1)h, and mn = P1"'Pk¢11"'QhIt follows that

a) = (—1)“ = 1. d|n Solution. This is clear for n = 1, when there is only one term in the sum

(namely p.(1)). For n > 1, if 191, . . . , pk are all the distinct prime factors of n, we realize after a moment of thinking, that

ZM(d)=#(1)+ Z #(p¢)+ Z #(Pipj)+“'+M(P1P2"‘Pk) d|n

ISiSk

ISi2

E10. Evaluate

i k!(k2 + 1). Ic=1

E11. Consider n arithmetic progressions with the same common difference d and having their first terms 1, 2, 3, . . . , n. If S'(n, k) is the sum of the first n terms of the arithmetic progression that has its first term k, prove that

S(n, 1) + S(n, 2) + S(n, 3) + - - - + S(n,n) = %2(2 + (n — 1)(d+ 1)). E12. Let a1,a2, . . . ,an be an arithmetic progression With ak aé 0 for all 1 S k S n. Prove that n—1

_n—1

k=1 akak+1— alan

1.

Easy Problems

173

E13. Let a1, a2, . . . , an be an arithmetic progression with positive terms. Prove that

1 + 1 + ,/a1 +1/a2 ‘/a2+‘/a3

+

1 _ n—1 ,/an_1+,/an ,/a1 +x/an'

E14. Prove the inequality

1 1 1 ——+—+---+—>24 fi+\/§

x/5+\/7

E15. Evaluate

1/9997+\/9999

1 2 _ _

i ’“ 1

16:1 [64+ +2

E16. Prove that

9:9

_1 1 —

k=1(\/E+ k+1) (VE+ 4k+1)

E17. Find the closed form

\/1+1+1+\/1+21—2+1 1 + 20002' 1 12 22 32 +-- -+ 1+ 19992 E18. Evaluate the sum

; 2k+¢4k2—1' 1 2

1 2

E19. Letan=\/1+(1+;) +\/1+(1—;) ,n21.Provethat 1 1 1 _+_+...+__ al 02 G20 is a positive integer.

174

Chapter 8. Problems

E20. Prove that

z”: k=0

1 l _ k.(n

_ 2" k) l

nl

E21. Evaluate n

n

n

1

2

n + 1'

E22. Prove that 72.

11—1

m

m— 1

+

n—2

+”_+

m—2

n—m

_

n+1

0

—

m

E23. Evaluate 2008

2008

2008

2008

(0-4 4 )+3( 5 )-4( 6)

2008 2008 — 2004(2006) + 2005 (2007).

E24. Evaluate the sum 61:: 21(3k_ 2k)(3k+1_ 2k+1) '

E25. Let Fn be the nth Fibonacci number (F1 = F2 = 1, and E, = Fn_1+Fn_2 for all n 2 3). Evaluate 00

: ————F'° k=2 Fk —1Fk+1. E26. Prove that for all n 2 3, fi k=2

1+k2+k+1

_

9

_32"‘—1

(k—1)3

1

n3—n 2

3 '

1.

Easy Problems

175

E27. Let 2' denote the imaginary unit. Evaluate

ll 1 + z’ + k(k + 1) 17.

k=1 1—i+k(k+1)' E28. Prove that the identity

(a:+y+z)5—(x5+y5+z5) = 5(x+y)(a:+z)(y+z)(m2+y2+z2+my+mz+yz) holds for any numbers :6, y, and 2:. E29. Evaluate the following sum for every positive integer n

7r

27r

(n — 1)7r

1+cos—+cos—+---+cos—. n n n E30. Evaluate

n Z(_1)k(k;-1) .

k=1 E31. Evaluate

n Z(_1)k1k2+1!

k=1

E32. Prove that 211.:

(_1)k+1

_

2n

H 12—22+32—---+(—1)’°+1k2 _ n+1‘ E33. If 7'1, 7'2, . . . ,7»; and t1,t2, . . . ,tn are real numbers, prove that n

n

2: 27'l COS(tk — t1) 2 0. 16:1 l=1

E34. Prove that

(x/E+tan1°) (x/E+tan2°)

(x/fi+tan29°) = 229.

176

Chapter 8. Problems

E35. Evaluate

(1 — cot 1°)(1 — cot 2°) - - - (1 — cot 44°). E36. Prove that


1, 0s

27r

cos

2”—1

47r

To

1

2n-1

271—1

2"-

E38. Let n be a given positive integer and let ak=200s2—n_k,k= 0,,.1..,n—1. Prove that (——_ 1)n— 1

:l:[0(

1—

ak)=1 + an

.

E39. The sequence {xn}n21 is defined by 1

2

x1 = 5, wk+1 = cck +£1.79.

Find the greatest integer less than 1

1

1

—+—+"'+ {131+1 IE2+1 $100+1

E40. Solve the problem left unsolved in the Introduction.

Namely, if n is

any given positive integer and f is defined by f (cc) = :L' — [EJ show by telescoping that

E [f 2 R be a function such that

()f( )1snotan1nteger fork=1,,.2 ..,p—l; (ii) f(k) + f(p — k) is an integer divisible by p, for k = 1,2, . . . ,p — 1. Prove that

p—

1

p—

1

f(“J 1 Z z— = - me) - P— k=1 1‘

p

p k=1

H26. If p > 3 is a prime number and :13, y, and z are integers such that :1; + y + z and my + :32 + yz are both divisible by p, then 33” + yp + 2p and

fly" + xpzp + 3/s are divisible by p2. H27. Let p be an odd prime and let

S —

1

4—2-3-4

+

1

+

5-6-7

—

+

1

q(q+1)(q+2)’

1

where q = 375. Assume that — — 28,, = g, for some integers m and n. Prove that m E n(mod p). H28. Let n be a positive integer, and let 2" be the highest power of 2 dividing n. Prove that 22’ is the highest power of 2 dividing the numerator of

1+1+1+ 3

5

+ 2n 1— 1

when the sum is represented as a fraction in its lowest terms. H29. Let n 2 2 be a positive integer, with divisors 1 = (11 < d2 < . . . < dk = n. Prove that d1d2+d2d3+- - -+dk_1d;c is always less than 17,2, and determine

when it is a divisor of n2. H30. Prove that

Md) _ __)(n

,ZT-‘b-n-

3. Ha'rd Problems

189

H31 . Prove that

n

Za(d)u (—) = n. d dln

H32 . Prove that 1,

Hdmd) = dl'n,

if n is not a power of a prime

1 1'f 11—17a , W1'th pprlme. ' 5,

H33 . Let an be a sequence of integers that satisfies Zad=2n for ala 1. dln

Prove that n [ an for all n 2 1. H34 . Prove that

00

¢ p’” — pl"—1 and the integers (11,. . . ,an. For any 0 S j S n denote by 5j and tj the number of sums of the form ail + + (11]., with 1 3 i1 < < ij 3 n which are, and,

respectively, which are not divisible by p (thus so = 1, to = 0). Prove that m .S' = Z(—1)j(n _ m +J)sm_j E 0

i=0 and

(mod 17’”)

3

m T = :(—1)j(n — m + ‘7)tm_j E 0

i=0

(mod pr).

3

H36 . Evaluate

11 1-2-3

g+ 12 5 2-3-4

42+ 13 5 3-4-5

53+ 5

190

Chapter 8. Problems

H37. Let a be a real number. Define the sequence (:L‘n)n21 recursively by $1 = 1 and mn+1 = a” +nxn, for n 2 1. Prove that

n = 0°

11.

":1

xn+1

H38. Evaluate

Z Z z 3+ 2)‘ i=1 j=1(i+3 H39. Prove that

00

4

:i—1

16:1 k4

90

H40. Evaluate

2— ‘— £2" 1 g k_1

k2 l + 1

H41. Let a1,a2, . . . , 0,100 be nonnegative real numbers such that

a¥+a§+---+a¥oo= 1Prove that

afiaz + agag + - - - + afiooal g ?. H42. Let x1, . . . , x100 be nonnegative real numbers such that xi+mi+1+xi+2g1foralli=1,...,100

(set x101 = 1131, x102 = .732). Find the maximal possible value of the sum 100 S = 2 mm”. i=1

3.

191

Hard Problems

H43. Prove that for any real numbers $1,122, . . . ,xn and any nonnegative real

numbers 7'1, 7'2, . . . ,rn the inequality 71.

Z min(r¢,rj)xi:1:j 2 0 M=1

holds. (The sum is over all pairs (i,j) with 1 S 2' S n and 1 S j _ ”

1

1

=;(H_(k+1)!) 1 =1_(n+1)!2 (17,277,)

Solution. Notice that

2(3n2 +1) = (n+1)3 — (n— 1)3

and

n3 —n=n(n— 1)(n+ 1),

therefore

N 3722 +1

2_(3n2 +1)

1”

1

1

”2— 2(713 - ”)3: 21112— (n3 — ")3 _ 5 ”2:22 (Mn — 1))3 _ ((n + 1)”)3) 21

=fi

1

2 ((N+1)N)3’

for any positive integer N Z 2. Consequently,

2— 3n2 +1 _ I'm

1 _1

n22 (n3—n)3 —N1—)oo

1—6

1

2 ((N+1)N)3

1 :1—6

198

Chapter .9. Solutions

E10. Evaluate

n

E k!(k2 + 1). [i=1

Solution. We have n

Zk!(k2+1) =Zn:k!((k+1)2—2(k+1)+2) k=1

k—l

R‘

p..-

H

W1%

((k + 1)!(k + 1) — 2(k + 1)! + 2k!) ((k+2)! — (k+ 1)!) —22n:((k+ 1)! — k!) k=1

= n+2)!—2! —2((n+1)!— 1!)

= (n+2)! — 2(n+ 1)! = n(n+ 1)!. We used (k + 1)!(k + 1) = (19+ 1)!((k + 2) — 1) = (k + 2)! — (k: + 1)!. E11. Consider it arithmetic progressions with the same common diflerence d

and having their first terms 1,2,3,. . . ,n. If S(n,k) is the sum of the first n terms of the arithmetic progression that has its first term k, prove that

S(n, 1) + S(n, 2) + S(n, 3) + - - - + S(n,n) = %2(2 + (n — 1)(d+ 1)). Solution. Indeed, we have that

S(, ,k)

n(2 k + (2n — 1 )d),

for each 1 S k: g n, therefore

n

n

_ n

gm ,k)—Zl(nk+

n(n—l)

2

—n "

d)_ gm

”n(n—l)

2

d

k:

_ n2(n+ 1) + n2(n — 1)

_

2

2

d=%2(2+(n—1)(d+1)).

1.

Solutions to Easy Problems

199

E12. Let a1,a2, . . . ,an be an arithmetic progression with ak yé 0 for all 1 S k S n. Prove that 11—1

1

72—1

2 k=1 a k ak+1 =a1an' Solution. Let d be the common difference of the progression, that is,

d=a2-a1 = "'=an—an—1,

and, more general, aj — a,- = (j — i)d for all i,j E {1, . . . ,n}.

The key observation (for telescoping) is that 1

akak+1

_1

d

d akak+1

1

d

ak+1-ak_1(1

akak+1

d

Gk

1)

ak+1

Therefore

:11 _1§(i_;)_1(i_i) d 01 an d k=1 ak ak+1

k_1 akak+1

_ 1

_d

an — a1 _ n — l

awn _ alan’

as desired. Many sums of this type are encountered; for instance,

:1— if L; _1_1_n_—1 k—(k+1) k=1 k k+1 _ n_ n’ 01‘

Z

1

_i"‘1

1

k=1(61~c—1)(6k+5)_6k=1 6—k—1

__1_ 1 1 _6 5 671—1

1 6—k+5

_ —1 _5(6n—1)'

'

200

Chapter 9. Solutions

This also allows the computation of the corresponding infinite series. Thus

1 1 =1 1 ——n) Z:Ic—(Ic14r1)= nlimoZk—(k+1): "Loco ’ n—l

and, similarly oo

.

1

;— (6k - 1)1(6k + 5) 13530; (6k — 1)(6k + 5) =1( _ lim 1

1

1

_n—>o06 5 Notice that a converse is also true. numbers a1, . . . , an, we have

“ 30'

Namely, if, for the nonzero real

1-1 k: 1

6n—1

_ 1

_1 akak+1=a1aj

for all 2 S j g n, then a1, . . . , an form an arithmetic progression. Indeed, by subtracting the above equation from the similar one obtained by replacing j to j + 1, that is from

we obtain

alaj+1

7

H

akak+1

j

j-1

ajaj+1

(hag-+1

alaj

After clearing the denominators and rearranging a bit, we get

(1' - 1)(aj+1 — aj) = a,- - a1 for 2 S j S n — 1. For example, when 3' = 2, this gives (13 — a2 = a2 — a1. We denote d = a3 — a2 = a2 — a1, and see that ak = a1 + (k — 1)d is true

1.

Solutions to Easy Problems

201

for k E {1, 2,3}. Assuming this to be true for k = j, we get (from the above equality)

(j - 1)(aj+1 — 02') = (J' - 1)d => aj+1 = 03' + d => aj+1 = a1 +jd, thus ak = a1 + (k — 1)d is shown to be true for all k S n, by induction — which is what we intended to prove. E13. Let a1, a2, . . . , an be an arithmetic progression with positive terms. Prove that 1

1

+

«01 + x/az

1

+ . . . +

V112 + «613

=

van—1 + y/a'n,

n— 1

_

V01 + van

Solution. Let d be the common difference of the progression, so that ak+1—ak=dforlnn—1. Then

11fak+1_

Gk _ln_1((WT— Wk)

Z— 1"“’“ +1,/ak+1=k:=1—ak+1 — ak

d k=1

+1

=(n_ )v—_—_ «‘1 _(_1)n_— r1 (n1)d

—

an—al

n—1

WWW; Prove in an analogous manner that n—l

Ic=1 fl+ ‘3/akak+1 + {Va—i:

W'l‘ v3 “Ian + {Va—31

E14. Prove the inequality 1

1

1

—+—+---+—>24 x/9997+\/9999 Vim/3 «Sh/7

202

Chapter .9. Solutions

Solution. We proceed as in the previous problem in order to get 5000

5000 «219—4—2— \/2k——1

Z1\/2k— 1+ 1v2k +1 :1;

= §(\/1000 — 1) > %, the inequality being equivalent to \/ 10001 > 100 4:) 10001 > 10000. Now, for the sum from the statement of the problem we have

250012500 2—11:( 1 1 > 1‘/4j— 3+x/4j— >2 J4j— 3+\/4j— 1 +\/4j— 1+\/4j+1 15000

9—9

=2 2 W> > 24’ as desired. E15. Evaluate

1 n

k2 — g

k=1 k4 + 1

Solution. We have 1 192—5

2 l _ (k+2 1 (k _l2)(k+k+2)

194+:

1 k 2_ k+2

(k2+k+%) (k2—k+%>

kz—k+%

k2+k+é

_ 19—; _ (mu—é 1924+; (k+1)2—(k+1)+%

1.

Solutions to Easy Problems

203

consequently

k——1

n

nk2—1

2:2 1 k=1k4+Z

k=1

(k+1)——1

21— k2—k+§

2

1

1

1 12—1+§

1 (n+1)2—(n+1)+§

71+1

:1.

1

(k+1)2—(k+1)+§

2

=

n2+n+l

n2+n+y

2

Note that this implies

”2 2

1

i

k2——

%=

k=1k4+— E16. Prove that 9999

1

k=1 («h m) (%+ e/m)

= 9.

Solution. Indeed, we have

(«c—Hm?) (%+%)(V—%)=(¢a+¢3)(f—¢B) =a—b for nonnegative a and b, thus

(M+fi)(M+mz)(w—mz)=1 forkZO, and

99f?

1

k=1(\/E+\/k+—1)(\4/E+\4/m)

=99f39(s/m—¢/t) k=l

= 410000—1=9.

204

Chapter 9. Solutions

E17. Find the closed form

\/1+i+i+\/1+i+i+ 12 22 22 32

+ 1+;+; 19992 20002'

Solution. Here we must observe that

1+1+ k2

1

_ k4+2k3+3k2+2k+1 _ (k2+k+1)2

(k+ 1)2 _

k2(k+ 1)2 1

_ k2(k+ 1)2

2

= (1+k(k+1))' Consequently, 1999

Z k=1

1

1 —

1 —=

+k2+(k+1)2

1999

1

1

Z

19:1

1999

=

1

1 1 ———

Z(+k k+1>

10:1

1 3999999 _ 1999 + 1 — 2000 _ 2000 ' E18. Evaluate the sum

Solution. Knowing the formulas

W=\/a+x/2mi\/a—x/2h2———b for de—nesting nested radicals of order 2 can be very helpful here.

(Of course these formulas work for nonnegative a and b such that a2—b 2 0 and they indeed de-nest the radicals only if a2 — b is a square.)

For a = 2k and b = 4102 — 1 (and with a plus sign) we find

Mme—rm: ”fin ”2‘1,

1.

Solutions to Easy Problems

205

and reveal the sum to be an old friend:

2W: figm+—2m

zgimc—m) k=1

= gU/WH' 1 — 1).

1 2

1 2

E19. Letan=\/1+ 1, too. This shows that

100

1

1

Z k=1 93k + 1 =

_ — $101 > 1

and completes the solution. E40. Solve the problem left unsolved in the Introduction. Namely, if n is any given positive integer and f is defined by (I;

W“) = m ‘ lil show by telescoping that °°

[’6]

E [f 2(n)J =n— 1’

k=0

where flk] is the kth iterate off (that is, flk] = f o f o - - - o f with k appearances of f; we also consider f[0] to be the identity function that

maps a: to x, for every at).

230

Chapter 9. Solutions

Solution. We saw in the Introduction that the numbers nk = f ““41 (n) eventually become equal to 1, hence the sum is actually finite (the cor-

responding terms in the sum are 0). Considering a: = f [k] (n) in the very definition of f yields

f(f[’°](n)) = f‘k](n) — [@J e [@J = flk](n) — 14““ (n). Thus, if ns+1 = 1 for some 3 2 0, then

°° 14%) 3‘1 Mn)

2 [—2 J—Z [—2 J 16:0

19:0 5—1

= Z (flklm) — flk+1](n)) k=0

= #010») — Mn) = n — 1, as desired.

2.

2

Solutions to Medium Problems

231

Solutions to Medium Problems

M1. For each positive integer k, let f(k) = 4’“ + 6’“ + 9k. Prove that for all

nonnegative integers m and n, f(2'm) divides f (2") whenever m is less than or equal to n. Solution. By repeatedly using the identity

(a2 — ab + b2)(a2 + ab + b2) = a4 + a2b2 + b4, we get

a2 ~(a2

n+1

m+1

+a2 b2 +b2 n

n

m

— a2 b2m + b2

n+1

m+1

= (a2

)(a2

m+1

m+2

+a2 b2 +b2 m

— a2

m+1

m

b2

m+1

m+1

+ b2

)

m+2

)m

‘ (a271, _ a2n—1b2n—l + b271,)

whenever m S n are nonnegative integers. This shows that, if a and b are integers, and m S n, then a2

m+1

m

+ a2 b2

m

+ b2

m+1

_

_

d1v1des a2

n+1

n

n

+ a2 b2 + b2

n+1

.

In particular, for a = 2 and b = 3 this means that f (2m) divides f (2"), as required. M2. Evaluate

12+22+32—42—52+62+72+82—92—102+---—20102, where each three consecutive + signs are followed by two — signs. Solution. We have

(5k — 4)2 + (5k — 3)2 + (5k — 2)2 — (5k — 1)2 — (5k)2 = 25k2 — 80k + 28, thus our sum is 402

2((5k — 4)2 + (5k — 3)2 + (5k — 2)2 — (5k — 1)2 — (519)?) k=1

232

Chapter 9. Solutions 402

402

402

402

= :(25192 — 8019+ 28) = 252k2 — 802k+2821 19:1

=2

5

16:1

k=1

k=1

W _ 80_402_' 403 + 28 . 402 = 536926141. 2

6

M3. Prove that

1+2q+3q2+---+nq"_1= l—nq” + q—q” ,

l—q

(l-q)2

for every q aé 1.

Solution 1. Brute force is our first approach: we multiply the given

sum by (1 — q)2 and just calculate. Thus

(1 — q)2(1 + 2g + 3g2 + - - - + nqn_1)

=(1-q)(1+2q+3q2+---+nq”‘1—q—2q2—3q3—----m1") =(l—q)(1+q+q2+---+q“‘1—nq'”) =(1-q)(1-nq”)+(1-q)(q+q2+---+q"‘1) =(1—Q)(1—nqn)+(q—qn)Now divide in this equality by (1 —— q)2 in order to obtain the desired result. Of course, here and elsewhere, the identity

(1-q)(1+q+q2+-~+qs‘1)=1—qs is taken for granted. Solution 2. Knowing Abel’s summation leads to this second approach. We use the identity TL

Zkak = ZOE]; + k=1

k=1

+an).

2.

Solutions to Medium Problems

‘

233

for CL], = q’“_1 and obtain

n k 1 —Zq n k 1 (1+q+ q —

k=1

—

—

nu.

n k 11_qn—k+1 +q k )—Zq 1_q n—

—

k=1

—

k=1

=1—ZqH—1—=1 1

n

_

nq"

—qk=1

—q

_1-nq'”

l—q

1—q“

nq”

( —4)

—q

q—q”

(1-q)2'

On the other hand, if we use 11

n

Zkak=n(al+”'+an)_Z(al+”'+ak—1) k=1

k=1

(that is, the same identity as above, but rearranged), we get n

1_

n

'n,

1_

751—1

1_

n

n

1_

16—1

qk_1= ”1—q ‘ k=1 1—q=“ 1 —qq ‘ k=2 1—q—q k=1 —q —q _n—nqn_n—1 1—q

1 — W" 1-q

= — +

q+q2+~-+qn’1

l—q

1—q

q — q” . (1 - q)2

Basically, we made the same calculations in each of these approaches — they differ only formally. Solution 3. For the reader with some basic knowledge in calculus, the following solution is available, too. We can consider the functions f and 9 defined, for 51: 7Q 1, by 1 _ $n+1

f(x)=1+x+---+a:”

and

g(:1:)=

1_x

We established that f(:1:) = 9(3)) for all real numbers as 76 1 (actually, even in the case x = 1 the equality remains true if we consider instead

234

Chapter 9. Solutions

of 9(1) the limit of g for a: tending to 1). Since f and g are equal, and both functions are differentiable, their derivatives must be equal:

f’(:1:) = g’(:c), that is

1+2x+---+mc‘"—

1 _ -(1"L+1):13"(1—:I:)+1—a:"+1

‘

(Ho)?

_1—n:r”

:r—sc"

l—x +(1—as)2 for every :1: aé 1. We used 2: instead of q in this solution, since it is a more common notation for a variable.

Finally we invite the reader to take a fourth (obvious) way for solving this exercise, namely to use induction. M4. Evaluate

z k=1 —’“—k4 + k2 + 1 Solution. The identity

(aZ—a+1)(a2+a+1)=a4+a2+1 comes to our mind (and becomes handy) again. We have n

1

1

:k—4+k2+1:k;§(k——k+1

k2+k+1)

n

ml; 2 =1 1 1_ =2

1

1k—+1

(k+1)2—(k+1)+1

1

_

(n+1)2— (n+1)+1

_2(n2+n+1)'

M5. Evaluate the sum

1 -2 —+ 3+1

32+1

+

22 34+1

+-~+

n2+n

2" 32"+1'

2.

Solutions to Medium Problems

235

Solution. As simple as it may seem, it is important to notice that

(32" — 1)(32’“ + 1) = 32"+1 — 1. Consequently

2k

2k(32’° — 1)

2k(32’“ + 1 — 2)

32" + 1 = 32"+1 — 1 = _ 219(32" + 1)

32"+1 — 1 2k+1

2k(32" + 1) — 2k+1

=

32"+1 — 1

2k

2k+1

32’“+1 — 1 _ 32"+1 — 1 _ 32" — 1 _ 32"+1 — 1

1

hence, by telescoping, i

2k

_

20

k=032k +1 _ 32° —1

_

2n+1

_ l _

2n+1

32"+1 —1'

32"+1 — 1 _ 2

M6. Let fn = 22" + 1, n = 1,2,3,... ve that

1+3+.+T*ngo

_

1+:z:+:172

1

_ 1+x+x2’

because, for a: 6 (—1,1), we have lim :5" = 0. 71—)00

M13. Let Fn be the nth Fibonacci number. Evaluate

35—1 k=2 Fk—l+1' Solution. Because Fk = Fk+1 — Fk_1, we have

1 _Fk+1_FIc—l_ 1 __ 1 Fk—l+1 Fk—lFk+1 Fk—l Fk+1’

2

;

242

Chapter 9. Solutions

and

f:= __1_ _ f (_1_ _ _1_) Fk—l+1 k=2 Fk—l FkFlc+1 — lim

n

—1

_n->°°k=2 Fk—l

——1

Fk+1

_nm( 1 _ FnFn+1 1): F1F2 1 =1. n—>oo F1F2 For the Lucas numbers Ln, defined by

L0 = 2, L1 = 1, and Ln = Ln_1 + Ln_2 for n 2 2 compute in the same way

°°

1

M14. Let n be a nonnegative integer. Prove that

$491425)Solution. Remember that

2.

243

Solutions to Medium Problems

The last step follows from the fact that

i< “ > 4, while for n = 1 it reduces to 2m — 2m/2 > 2, hence to

2m/2 > 2. Note that for m = 2 the above inequality becomes

n6—n4—4n3+4>0%(n—1)(n5+n4—4n2—4n—4) >0 and it still holds for n > 1. So, the given sum is not a prime for m 2 2 even and any positive integer n with the only exception of m = 2 and n = 1 when the sum equals 13.

M30. Let p be a prime such that p E 1 (mod 3) and let q = [213/3]. If

m+m+ +44 1-2

3-4

(q—l)q_n

for some integers m and n, prove that p | m. Solution. Suppose p = 3s + 1, with s a positive integer. Then

q = [210/3] = L28 + 2/3] = 23, and the sum from the statement of the problem actually is _1_+_1_+...+___1__=1_l+l_l+...+;_i

1-2

3-4

(2s—1)2s

_ 1

2

3

1

4

23—1

1

_s+1+s+2+”.+%‘

23

2.

Solutions to Medium Problems

261

23

So, we have to prove that if we write the sum 2 — as an ordinary k=s+1

fraction, then the numerator of this fraction is divisible by p. But we have 23

1

1

1

=5 2 (Wm) =5 2 —k(3s+i_k> 1 1

k=s+1 23

k=s+1

p

2 k=s+1 k(3s + 1 —— k) Thus we have written the sum as another sum in which the numerator of each term is p, and the factors from the denominators are strictly less

than p (including the factor 2). This means that none of these fractions can be simplified by p, therefore the factor p remains after addition is performed, and cannot be simplified, hence the final numerator is divisible by p, as we intended to prove. For example, when p = 13 we have q = 8 (s = 4), and the sum from the statement of the problem is

1 + 11.1.,1 _1_1+1_1 1_1+1_1

1-23-45-67-8_2345678

_1+1+1+1

_5678 _ 1 13 + 1 3 + 13 + 13 _2

5~8

6-7

7-6

8-5

'

For the sake of completeness we prove the middle equality (but we proved it before in Example 4.6 — and we suppose that the careful reader already did it again), that is, in the general case,

11+11++1 2 3 4 23—1

1—1+1++i 23_s+1 s+2 28'

262

Chapter 9. Solutions

Indeed we have

11_111

— 1+1+ 2

111

+ls -i+ 1 + —s+1 5+2

11

+i 23'

M31. Prove that for different choices of the signs + and — the expression

:|:1:|:2:l:---i(4n+1) yields all odd positive integers less than or equal to (2n + 1)(4n + 1). Solution. This is clearly true for n = 0, so we will assume it true for n — 1 and prove that it also holds for n. Thus, the induction hypothesis says that the sums

i1:l:2:l:---i(4n—3) produce (for various choices of the i signs) all the odd positive integers

at most equal to (2n — 1) (4n — 3). Therefore, these numbers can be also achieved as sums of the form

:l:1:l:2:|:~--:l:(4n—3)+(4n—2)—(4n—1)—4n+(4n+1) as required.

In what concerns the other odd positive integer until (2n + 1)(4n + 1), we first note that 1+2+---+(4n+1)=(2n+1)(4n+1). We can subtract 23' (with 1 S j 3 4n + 1) from this sum and get a representation of (271. + 1)(4n + 1) — 2j, namely

(2n+1)(4n+1)—2j=1+2+---+(j—1)—j+(j+1)+---+(4n+1).

2.

Solutions to Medium Problems

263

We have thus represented all the odd numbers (in decreasing order) from

(2n + 1)(4n + 1) until (2n + 1)(4n + 1) — 2(4n + 1) = (2n — 1)(4n + 1). Now, by the same idea, we subtract from

1+2+'--+4n— (4n+1) = (2n—1)(4n+1) all the numbers 2k with 1 S k 5 4n, and get representatiOns for all the

odd numbers

(2n— 1)(4n+ 1) — 2112,16} = 1,2,...,4n. Thus we have representations for all the odd positive integers from

(2n + 1)(4n + 1) until (2n — 1)(4n + 1) — 8n = 877,2 — 10n — 1. On the other hand, as we have seen in the beginning, all odd positive

integers from 1 to (2n — 1)(4n — 3) = 8n2 — 10n + 3 also do have representations. As 8n2 — 10n + 3 > 8n2 — 10n — 1 the problem is solved. M32. Let n be a positive integer. Prove that all binomial coefi'icients (2) with 0 S k _0Also, of; — bi = inky], holds for every 1 S k S n. Thus, by using Abel’s summation formula and all the above observations, we have

kzak- 217},'= 2231411; = 961(111 - 112) + ($1 + 932x311 - 112) :1

k: 1

k=1 +"'+(m1+x2+"l+wn_1)(yn_l_yn)

+(x1+x2+---+xn)yn20 and the conclusion follows for every j 2 1. M35. Evaluate

if: 1 k=1 1:]: l'

Solution. Remember that, for any :1: E R, we have °°

m

Z 96—. = 6””,

m=0 m.

the well—known formula that defines the exponential, in which the absolute convergence of the series is also well-known. Thus, in particular,

°° 1

Z W' = 6’ m=0 where the series is absolutely convergent. Consequently, for the series

from our problem the (absolute) convergence is clear, too, and reversing the order of summation is allowed. And we have

2271. =ZI%Z =2; SHEET. k:1 l>kl l=11 1 we have ’n — 1

'n,—1

lim _2_"’9__= lim L20, n—)ooa,n+a,'"'+1

11,—)00

1

1+—n+ a

1 a

2n

the right-hand side of the previous identity also has limit 0 when n —> oo.

282

Chapter 9. Solutions

Consequently, the required sum is

Z(-1)

k_alzzk::12k:—111

— =n151302(-1)

k_—12k1a2k-—11

-———+1

=a2+1a+1

H4. For a nonnegative integer k, define Sk(n) = 1'c + 2’“ + - - ~ + nk. Provethat

1

1"

1+

8' n = ( n+1’". 1266)“) )

Solution. This is just a reminder (we did it before at the start of Chapter 1). We have

2(Z)Sk=§(2)i =i"l(;)r k=0

j—l

j==1k0

k

=Z((j+1)’—j’) =(n+1)r—1, i=1

as required. The following recurrence for the sums Sk (n) also holds:

Sr+1(n) - UlSrUl) + USSr—fln) - - - - + (—1)'0$51(n) _ (n+1)n(n—1)---(n—r) — 7‘ + 2 ’ for positive integers 1' and n, Where a; is the sum of all possible products of k distinct factors chosen from the numbers 1,2, . . . ,r (the kth fundamental symmetric sum of the numbers 1, 2, . . . ,r), for 1 g k S 7'. Indeed,

Sr+1(n) — 01.5201) + 038' -1(n) — - ~ ~ + (—1)T0,’:81(n) n

= Zr“ — (7;e + 0521'“ — ---+ (—IYZJ' i=1

i=1

i=1

i=1

3.

Solutions to Hard Problems

283

:20?“ (”+071-+(-10rj)=zj(j_1)'”(j_r)

—:§Z((j+1)j~-(j—r)—j -(j—r)(j—r—1)) = (n+1)n(n—1)-~(n—r)l r+2

Get a slightly different recurrence by using the similar formula

j’+l+ai’j“+0n 1+---+a:j=j(j+1)---(j+r)For instance, when r = 2, we have 0% = 1 + 2 = 3 and 0% = 1 - 2 = 2, therefore we get 53(7),) — 352 (n) + 251(7),) =

(n + 1)n(n — 1)(n — 2)4

By replacing here Sl(n) = n(n + 1)/2 and 32(7),) = n(n + 1)(2n + 1)/6, we get S3 (n) = (n(n + 1) /2)2, as we know. Use this recurrence formula to find S4(n). H5. Find all positive integers n such that m

n = H(ai + 1), i=0

where amam_1 . . . a0 is the decimal representation of n. Solution. For m = 0 there clearly are no solutions, so that we assume

m 2 1 (that is, n has at least two digits). We have, for a solution n = amam_1 . . . a0,

W407“ 3m: (am+1)(am_1 +1)---(ao+1) 5 (am + 1)(am_1 + 1) - 10m—1.

284

Chapter 9. Solutions The first inequality comes from the fact that all digits are at least 0, and

the second one is due to the fact that the digits are at most 9 (we use these bounds for the last m — 2 digits). Consequently

10am + am—1 = (Imam—1 S (am + 1)(am—1 + 1), which is equivalent to

(9 — am—1)am S 1. Since both 9 — am_1 and am are from the set {0,1, . . . ,9}, the last inequality is possible for either am = 1 and am_1 = 8, or am_1 = 9 and am any other digit. In the first case the above inequalities become

1810"“1 g 18a'm12‘. . . a0 = 2 . 9. (a,,,_2 + 1) - - - (a0 + 1) g 18-10m_1. That is, they must be all equalities, which is possible if and only if am_2, . . . ,ao are all 0, and, simultaneously, they are all equal to 9. Of course, this can only happen when m = 1 (so there are no am_2, . . . , a0) — that is we get the solution n = 18. In the second case the equality

m=(am+1)~-(ao+1) forces (10 to be 0 (because there is a factor of 10 in the right-hand side), which makes the right-hand side at most equal to 10’”. On the other hand, the left hand side is at least 19 - 10m_1, as am 2 1, and am_1 = 9. Since 19- 10m—1 > 10”” the equality is not possible in this case. The only solution remains the number 18. H6. Let k ah:

_ (k—1)%+k%+(k+1)%'

Prove that (11 +a2+-"+a999 < 50.

3.

Solutions to Hard Problems

285

Solution. We have 999

999

’6

k=1

k=1(k_1)%+k%+(k+1)§ 999 k

2w:

n= (%)n—(\/m—a)"< (37131 and, also,

—(a— a2+1)n+ G)”: (1)”? (x/tfl—H-a)n>02 2 n

]_

Thus in both cases the expression — (a. — Va? + 1) + (5) 0 and 1, therefore its floor function is 0. So,

an: (a+\/a2+1)n+ (a—x/a2+1)n

n

is between

3.

Solutions to Hard Problems

287

for every n 2 0. Because (a + Va2 + 1)” and (a — Va2 + 1)” are the roots of the quadratic equation x2 — 2am — 1 = 0, it follows that the numbers an satisfy the recurrence an+1 — 2aan — an_1 = 0 4:) an+1 — an_1 = 2a¢1n

for n 2 1, hence 00

00

00

:————i:— 2a "I: an _1anan+1 =lzfliifl 2a ”:1 an—lanan+l

n_1 an_1an+1

_ 1 7: _ 2a.

1 an_1a,n

1

1 _ 1

anan+1=2aa0a1 _ 8a2’

because a0 = 2 and a1 = 2a. H8. Evaluate

00

zimi 16:1 2 k 2k 1 where a 75 s7r, with 3 any integer.

Solution. Remember (and prove!) the formula tanx = cota: — 200t2a3, according to which we can write 2— tan: =2

2’“

19-1

1 cot— —ico——t

27

a

—.cot—a——cota 1

2k2k—12k 1:271

2n

Consequently, 1 1 Eflktan2—=limoo(2—ncot§1;—cota)=E—cota.

This is because we have the well-known limit lim xcotx— — lim

tit—)0

:1:

z—)0 tan a:

=1

288

Chapter 9. Solutions

yielding l'

1

ta-l'

1

a

”320mm 2n—nfifioa

ta_

2nco 2n_a

(as a/2" —> 0 when n —> 00). H9. Let n be a positive integer. Prove that n—l

2

—1 k 1 2 2 1_I(2sin2 (lg—)1 + 23m2 (—ifl — sin2 I) = (1 — cos” I) . n n n 72

19:0

Solution. As we know, we have the factorization n—l

z”— 1 = H (z—cos2k—7r —isin2k—7r> n n k=0 for any complex number 2. Taking the absolute values, we get n—l

2k n

2

|zn— 1|2 = H z—coszk—7r —isin—7r 16:0

hence

n

n_

,

I

2k

Iz” — 1|2 = H (2:2 — cos—fl- + 1), k=0

n

when 2 is a real number.

For 2 = cos 27" we get

2

(1 — cos” I) n

2

=

”‘1 k=0 11—1

=

16:0

2

2

2k

(cos2 _7r — 2cos _1r cos —7r + 1) n n 77. 2 n

(cos2—7r—cos

2 k—1 2 k 1 ( )fl—cos ( + )W+1) n n

n—l

— 1 = H (23in2 M + 2sin2 M — sin2 21) [6:0

n

n

after a few simple trigonometric transforms, as desired.

n

3.

Solutions to Hard Problems

289

H10. Let m and n be integers greater than 1. Prove that

k1+l€2+-..

, 11k2,...,kn>0

+kn=mk

1.

n. so

2,

—-k'k'--k'

k S(( 1+2k2+---+nkn) n)

0

Solution. The sum is over all n—tuples (k1, k2, . . . , kn) of nonnegative

integers that sum to m and we first observe that it represents the real part of ——1——6 i(k1+2k2+m+nkn) 2?" k1+k2+"'+k‘n=mykl)k2v-"1kn20

k1 !k2 !---kn ! 1

L13 k

.12 k

.1; kn

k1!k2!---kn!(82" ) 1(“32 2") 2"‘(en 2") Z = k1+k2+~--+kn=m,k1,k2,...,kn20 1 = _(1+w+n_+wn—1)m

m! for w = e%. Thus we arrived at the sum of the nth roots of unity, which is well-known to be 0, hence the conclusion follows (and we see that it remains true if the cosines are replaced by Sines, too). We used the multinomial formula

m! _m’f1x’

E k1+k2+...+kn=m,k1,k2,...,kn20

k1!k2!---kn l

§2m xk"=(x1+x2+-- --+:rn)m

H11. Let X be a set with n elements. Prove that

Z |YnZ| =n-4”_1. nx

The sum is over all possible pairs (Y, Z) of subsets of X. Solution 1. First we note that if M is a finite set with m elements, then the number of pairs (A, B) of disjoint subsets of M is 3’”. Indeed,

we can choose A having k S m elements (from the m elements of M)

in (7:) ways. Once we chose A, B can be any subset of the complement

290

Chapter 9. Solutions

M \ A of A with respect to M. Since M \ A has m — k elements, there are 2m_k possibilities for B (for every subset A of M with k elements). Thus the number of such pairs (A, B) is

f: (7:) 2"” = (2 + 1)m = 3’". k=0

Now we evaluate the sum from the statement of our problem in the following way. For a given subset S of X, we can have Y 0 Z = S for

Y, Z Q X in as many ways as there are pairs of disjoint subsets of X \ S. (This is because Y 0 Z = S is possible if and only if Y = S U Y1 and Z = S U Z1 for mutually disjoint Yl and Z1 with elements outside S.)

According to the observation from the beginning, we see that there are 3IXHSI pairs (Y, Z) of subsets of X such that Y 0 Z = S. Of course, if n S has k S n elements, it can be chosen in (k) ways, and it contributes

to the sum with an amount of Is. All these being said, we can calculate

"

n

n

n

_ Zk+z(nbfll) k>0

k>0

k>0

kZI

k>1

[6—1

n—ld‘S —2

for all n > 2. Since 80— F1, 8'1: F2, and the sequences (Sn)n>0 and (Fn+1)n>o satisfy the same recurrence relation, they must coincide, that is, 51,— — n+1 for every n > 0, as required. H18. Evaluate

(z)—(”;1)+(”;2)-(”;3)+~~ Solution. That is, we are asked to evaluate

Z(—1)’“(”;’“),

kZO

300

Chapter 9. Solutions

which suggests us to consider the more general problem of evaluating

the sum

snug) = Z (n ; k):1:k. kzo With this notation, we must determine Sn(—1) (after we found, in the previous problem, that Sn(1) = Fn+1). A recurrence relation can be obtained exactly as in the preceding exercise:

511(93)=Z(n—:—1)xk+2(n;f;1)xk 1:20 =2

kZI

n—l—k) :1: k +532 (n—2—(k—1)> 3: ,4

k20
0( E —1 ) k _— Sn ( —1 ) = cos —+— 3 fism — 3 = — ‘/_cos —. 6 A period of 6 is immediately detected, based on the periodicity of the

trigonometric functions.

More precisely Sn+6(—1) = Sn(—1) for all

n 2 0. The sequence starts 1, 1,0, —1, —1,0, then these values repeat

indefinitely. For instance, Ssm(—1) = 1 for every nonnegative integer m.

3.

Solutions to Hard Problems

301

H19. Partition the set of positive integers into n 2 1 arithmetic progressions with first terms a1, a2, . . . , an and common differences d1, d2, . . . ,dn respectively. Prove that n ak_n_+1'

kg?‘

2

Solution. Note that we must have oh 3 dk. To see this, suppose the contrary ak > dk. Then x = ak — dk must be contained in some arithmetic progression, say the i—th, and we must have i aé k since at is below the start of the k—th arithmetic progression. But then CB + did]; = (1k + (d; - 1)dk

is in both the i-th and k-th arithmetic progressions, a contradiction. From this it is not hard to see that among the first N = d1d2 - - -dn positive integers there are precisely N/dk numbers belonging to the pro— gression Pk with first term a], and common difference dk. This is why we have

n'

.Mz

+ IV

E

M 2

N

3:1

N

n

1

._ —= I+dn©2dk

+—— +

n

M:

N

N=_

Z

k=1 lgjgN, jePk

k=1

i L, M L dk _ dkk 2oz,c dk k=1

'n.

=N

ak _

N

211,

—

N/dk—l

j=z Z (ak+idk)

1

Nn

———

i=0

302

Chapter .9. Solutions

by using the hypothesis and the observations from the beginning. This can be rearranged as

and the problem is solved. H20. Let a1 3 a2 3 that

3 an and b1,b2, . . . ,bn be positive real numbers such

a1+a2+---+ak2b1+b2+m+bkforall13kgm

Prove that alaz - - . an 2 blb2 . . . b". Solution. We have Valaz...an2

1

b

b

b

a1

0'2

an

Valaz...anlr_b (._1+_2+...+_n)

b1 ()2

>"a1a2

b

ann__..._1l a1 (12 an

= {71)l . . . bn,

whence the required inequality immediately follows. The second inequal-

ity that we used is, of course, the AM—GM inequality, so we still need to explain the first, that is, to prove

b1 b bn —+2+-- +—