ADVANCED PROBLEMS AND SOLUTIONS

EDITED BY
FLORIAN LUCA

Please send all communications concerning ADVANCED PROBLEMS AND SOLUTIONS

to FLORIAN LUCA, SCHOOL OF MATHEMATICS, UNIVERSITY OF THE WITWA-
TERSRAND, PRIVATE BAG 3, WITS 2050, JOHANNESBURG, SOUTH AFRICA or by
e-mail at the address florian.luca@wits.ac.za as files of the type tex, dvi, ps, doc, html, pdf,
etc. This department especially welcomes problems believed to be new or extending old results.
Proposers should submit solutions or other information that will assist the editor. To facilitate
their consideration, all solutions sent by regular mail should be submitted on separate signed
sheets within two months after publication of the problems.

PROBLEMS PROPOSED IN THIS ISSUE

H-842 Proposed by Hideyuki Ohtsuka, Saitama, Japan

Given an integer n ≥ 0, find a closed form expression for the sum
X
Fa+b Fb+c Fc+a .
a+b+c=n
a,b,c≥0

H-843 Proposed by Hideyuki Ohtsuka, Saitama, Japan

If integers a and b have the same parity with a > b > 0 and c is odd, show that
(Fa − Fb ) | (Fac − Fbc ) and (La − Lb ) | (Lac − Lbc ).

H-844 Proposed by Robert Frontczak, Stuttgart, Germany

Let Bn = Bn (α, β) be a generalized balancing number given by B0 (α, β) = α, B1 (α, β) = β
and for n ≥ 2,
Bn (α, β) = 6Bn−1 (α, β) − Bn−2 (α, β).
Prove that
2n  
X 4n
B2k (α, β) = (26n−1 + 24n−1 )B2n (α, β)
2k
k=0
and
⌊(4n−1)/2⌋  
X 4n
B2k (α, β) = (26n−1 − 24n−1 )B2n (α, β).
2k + 1
k=0

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THE FIBONACCI QUARTERLY

H-845 Proposed by D. M. Bătineţu, Bucharest, Romania, and N. Stanciu, Buzău,

Romania
Compute
   
lim lim (f (x + 1))Ln /((x+1)Fn+1 ) − (f (x))Ln /(xLn+1 ) xLn−1 /Ln+1 ,
n→∞ x→∞
where f : R+ → R∗+ is a
∗ function that satisfies limx→∞ f (x + 1)/(xf (x)) = a ∈ R∗+ .

SOLUTIONS

Evaluating the sum of a series of reciprocals

H-810 Proposed by Ángel Plaza, Gran Canaria, Spain

(Vol. 55, No. 3, August 2017)
Prove that
∞
X 1 5 1
4
= − √ .
Ln − 25 63 6 5
n=3

Solution by the proposer

Ln+2 + Ln−2
First note that L4n − 25 = Ln−2 Ln−1 Ln+1 Ln+2 and also that Ln = . Therefore,
3
1 1/3 1/3
= + . (1)
L4n
− 25 Ln−2 Ln−1 Ln Ln+1 Ln−1 Ln Ln+1 Ln+2
Taking into account relation (23) in [1]:
n−1  
X 1 1 1 Fn−1 3Fn Fn+1
=− + + +
Li Li+1 Li+2 Li+3 8 10 Ln Ln+1 Ln+2
i=1
and letting n approach infinity, we get
∞
X 1 1  √ 
= 5−2 5 .
L L L L
n=1 n n+1 n+2 n+3
40
Therefore,
∞
X 1/3 1  √ 
= 5−2 5 ,
L L L L
n=3 n−2 n−1 n n+1
120

from where the desired sum follows via (1). 

[1] R. S. Melham, Finite sums that involve reciprocal of products of generalized Fibonacci
numbers, Integers, 13 (2013), A40.
Also solved by Brian Bradie, Dmitry Fleischman, Hideyuki Ohtsuka, and Raphael
Schumacher.

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 ADVANCED PROBLEMS AND SOLUTIONS

Evaluating the sum of another series of reciprocals

H-811 Proposed by Ángel Plaza, Gran Canaria, Spain

(Vol. 55, No. 3, August 2017)
For any positive integer k, let {Fk,n }n≥0 be defined by Fk,n+2 = kFk,n+1 + Fk,n for n ≥ 0
with Fk,0 = 0, Fk,1 = 1. Prove that
∞ √
X 1 k2 + 4
= .
n=0
1 + Fk,2n+1 2k

Solution by Brian Bradie

√
Let k be a positive integer, and put αk := (k + k2 + 4)/2. Then,
!
1 n
   
1 n 1 2n+1 1
Fk,n = √ αk − − and Fk,2n+1 = √ αk + 2n+1 .
k2 + 4 αk k2 + 4 αk
Moreover,
√
1 k2 + 4α2n+1
k
= √
1 + Fk,2n+1 α4n+2
k + k 2 + 4α2n+1 + 1
k
√ 1
!
2
k +4 αk αk
= − 2n+1 .
k α2n+1
k + αk αk + 1
αk
Since
1
αk α
2n+3 = 2n+1k 1 ,
αk + αk αk + αk
it follows that the desired series telescopes and so its value is
∞ √ √
X 1 k2 + 4 αk k2 + 4
= · = .
1 + Fk,2n+1 k αk + αk 2k
n=0

Also solved by Dmitry Fleischman, Hideyuki Ohtsuka, Raphael

Schumacher, and the proposer.

An identity with sums of products of binomial coefficients

H-812 Proposed by Hideyuki Ohtsuka, Saitama, Japan

(Vol. 55, No. 3, August 2017)
Prove that
X Fn+1 Fn Ln  X Fn Ln  2j 
= .
i i j i i Fn+1
i+j=Fn+1 i+j=Fn+1

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THE FIBONACCI QUARTERLY

Solution by the proposer

Let a = Fn+1 , b = Fn , and c = Ln . Note that b + c = 2a. We have

    
b c i j b! c! i! j!
= · · ·
i j k k i!(b − i)! j!(c − j)! k!(i − k)! k!(j − k)!
b! c! (b − k)! (c − k)!
= · · ·
k!(b − k)! k!(c − k)! (b − i)!(i − k)! (c − j)!(j − k)!
    
b c b−k c−k
= .
k k b−i c−j
Using Vandermonde’s identity and the above identity, we have
X abc X bc X a   
i j
=
i i j i j k k
i+j=a i+j=a k=0
a    X   
X b c b−k c−k
=
k k b−i c−j
k=0 i+j=a
a    
X b c b + c − 2k
=
k k b+c−a
k=0
a     X bc2j 
X b c 2a − 2k
= = .
k k a i i a
k=0 i+j=a

A cyclic inequality

H-813 Proposed by D. M. Bătineţu-Giurgiu, Bucharest, and Neculai Stanciu,

Buzău, Romania (Vol. 55, No. 4, November 2017)
If xk > 0 for k = 1, . . . , n and m ≥ 0 is an integer, prove that
n
!
X 1 X x1 x2 x3 n2
≥
xk Lm x2 x3 + Lm+1 x3 x1 + Lm+2 x1 x2 2Lm+2
k=1 cyclic

and that the same inequality holds with the Lucas numbers replaced by the Fibonacci numbers.

Solution by Wei-Kai Lai and John Risher

We will prove that for positive A, B, C such that A + B = C, the inequality

 
n n
!
X 1 X xi xi+1 xi+2 n2
≥

xk Axi+1 xi+2 + Bxi+2 xi + Cxi xi+1 2C

k=1 i=1
cyclic

286 VOLUME 57, NUMBER 3

 ADVANCED PROBLEMS AND SOLUTIONS

holds. According to the AM-HM inequality,

n
X xi xi+1 xi+2 n2
≥  
Axi+1 xi+2 + Bxi+2 xi + Cxi xi+1 Pn A
+ B
+ C
i=1 i=1 xi xi+1 xi+2
cyclic cyclic
n2 n2
= P  = P  .
n 1 n 1
i=1 xi (A + B + C) i=1 xi (2C)
Therefore,  
n n
!
X 1 X xi xi+1 xi+2 n2
≥
 
xk Axi+1 xi+2 + Bxi+2 xi + Cxi xi+1 2C

k=1 i=1
cyclic
as claimed, thus proving the two inequalities in the original problem.
Also solved by Dmitry Fleischman and the proposers.

Errata: For Advanced Problem H-838 (Vol. 57, No. 2, February 2019) “Ln−(n+j) ”
should be “Ln−(r+j) ”. Furthermore, at the beginning of the published solution to Advanced
Problem H-809 (Vol. 57, No. 2, February 2019) in the formulas for p2m and p2m+1 in the
right sides, the exponents of L should be “2k − 1”, “2k”, and “2m + 1” instead of “k − 1”,
“k”, and “m + 1”, respectively. The editor apologizes for these inconveniences.

AUGUST 2019 287