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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 WITWATERSRAND, PRIVATE BAG X3, WITS 2050, JOHANNESBURG, SOUTH AFRICA or by
e-mail at the address [email protected] 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-773 Proposed by H. Ohtsuka, Saitama, Japan.
Let Bn be the Bernoulli numbers defined by the generating function
∞
X
x
Bn n
=
x .
x
e −1
n!
n=0
For integers n ≥ 0 and m ≥ 0, prove that
" L
#
n m
X
X
2n
n
m
2n−1
F2mk B2(n−k) = √ 2
(α − r)
+ Lm(2n−1) .
2k
5
r=1
k=0
H-774 Proposed by G. C. Greubel, Newport News, VA.
1. Let m ≥ 0, p ≥ 0 be integers. Evaluate the series
∞
X
Fn+p Ln+m
n=0
(n + p)!(n + m)!
in terms of the Bessel functions.
2. Evaluate the case m = p in terms of a series of modified Bessel functions of the first
kind. Take the limiting case m → 0.
3. Show that when p = 0 the series is given by
∞
X
Fn Ln+m
1
= √ (Im (2α) − Im (2β)) − Fm Jm (2).
n!(n
+
m)!
5
n=0
H-775 Proposed by H. Ohtsuka, Saitama, Japan.
Let c be any real number c 6= 2, −L2n for n ≥ 0. Let
∞ √ Y
c −1
γc = 5
1+
.
n
L
2
n=1
AUGUST 2015
279
THE FIBONACCI QUARTERLY
Prove that
∞
X
k=1
1
γc + c − 3
= 2
.
(L2 + c)(L4 + c) · · · (L2k + c)
c −c−2
H-776 Proposed by H. Ohtsuka, Saitama, Japan.
Determine
∞
X
(i)
n
(−1) tan
−1
n=0
1
L3n
and
∞
X
(ii)
tan−1
n=1
1
1
tan−1
.
F2n
L2n
SOLUTIONS
Sums of Fibonacci Numbers with Indices Given by Quadratic Forms
H-742 Proposed by H. Ohtsuka, Saitama, Japan.
(Vol. 51, No. 3, August 2013)
For positive integers n, m and p with p < m find a closed form expression for
n
X
k1 ,...,km =1
F2k1 · · · F2km F2(k12 +···+kp2 −kp+1
2
2 ).
−···−km
Solution by the proposer.
We have
!p n
!m−p
n
X
X
2k 2
2k 2
α F2k
β F2k
=
k=1
k=1
n
X
α2(k1 +···+kp −kp+1 −···−km ) F2k1 · · · F2km ,
n
X
β 2(k1 +···+kp −kp+1 −···−km ) F2k1 · · · F2km .
k1 ,...,km =1
2
2
2
2
2
2
2
2
(1)
and
n
X
β
2k 2
F2k
k=1
!p
n
X
2k 2
α
F2k
k=1
!m−p
=
k1 ,...,km =1
(2)
We have
n
X
k=1
2k 2
α
F2k =
n
X
2k 2
α
k=1
α2k − α−2k
√
5
n
1 X 2k(k+1)
=√
(α
− α2(k−1)k )
5 k=1
1
αn(n+1) − α−n(n+1)
√
= √ (α2n(n+1) − 1) = αn(n+1) ·
5
5
n(n+1)
=α
Fn(n+1) .
(3)
Similarly,
n
X
2
β 2k F2k = β n(n+1) Fn(n+1) .
(4)
k=1
280
VOLUME 53, NUMBER 3
ADVANCED PROBLEMS AND SOLUTIONS
Using (1), (2), (3) and (4), we have
n
X
k1 ,...,km =1
1
=√
5
(
F2k1 F2k2 · · · F2km F2(k12 +···+kp2 −kp+1
2
2 )
−···−km
n
X
k=1
2k 2
α
F2k
!p
n
X
β
2k 2
F2k
k=1
!m−p
−
n
X
β
2k 2
F2k
k=1
!p
n
X
2k 2
α
k=1
F2k
!m−p )
1 p
m−p
p
m−p
= √ αpn(n+1) Fn(n+1)
β (m−p)n(n+1) Fn(n+1)
− β pn(n+1) Fn(n+1)
α(m−p)n(n+1) Fn(n+1)
5
1
m
m
= √ (α(2p−m)n(n+1) − β (2p−m)n(n+1) )Fn(n+1)
= F(2p−m)n(n+1) Fn(n+1)
.
5
Also solved by Dmitry Fleischman.
On the Fermat Quotient Modulo p
H-743 Proposed by Romeo Meštrović, Kotor, Montenegro.
(Vol. 51, No. 4, November 2013)
Let p ≥ 5 be a prime and qp (2) = (2p−1 − 1)/p be the Fermat quotient of p to base 2. Prove
that
(p−1)/2
1 X (−3)k
qp (2) ≡ −
(mod p).
2
k
k=1
Solution by the proposer.
Since
√
π
π
1 ± i 3 = 2 cos ± i sin
,
3
3
applying the de Moivre’s formula, we have
√
√
π
π p π
π p + cos − i sin
(1 + i 3)p + (1 − i 3)p = 2p cos + i sin
3
3
3
3
pπ
p
p
= 2 · 2 cos
=2 ,
3
(1)
where we used the fact that p ≥ 5 is odd and so cos(pπ/3) = 1/2.
On the other hand, by the binomial theorem, we obtain
p p √ p
√ p X
√
p √ k X p
(1 + i 3) + (1 − i 3) =
(i 3) +
(−1)k (i 3)k
k
k
k=0
=2
k=0
X
0≤k≤p−1
2|k
=2
(p−1)/2 X
k=1
AUGUST 2015
(p−1)/2 X
√
p √ k
p
(i 3) = 2
(i 3)2k
k
2k
k=0
p
(−3)k + 2.
2k
(2)
281
THE FIBONACCI QUARTERLY
The equalities (1) and (2) obviously yield the identity
(p−1)/2 X
k=1
By the identity
p
2k
p
(−3)k = 2p−1 − 1.
2k
(3)
p p−1
=
with k = 1, . . . , (p − 1)/2, (3) becomes
2k 2k − 1
(p−1)/2
1 X (−3)k p − 1
2p−1 − 1
=
:= qp (2).
2
k
2k − 1
p
(4)
k=1
Finally, since
p−1
2k − 1
(p − 1)(p − 2) · · · (p − (2k − 1))
(2k − 1)!
(−1)(−2) · · · (−(2k − 1))
≡
(mod p)
(2k − 1)!
=
≡ (−1)2k−1
(mod p) ≡ −1 (mod p),
substituting this into (4), we obtain the desired congruence.
Inequalities Involving Sums of Reciprocals of Fibonacci and Lucas Numbers
H-744 Proposed by D. M. Bătineţu-Giurgiu, Bucharest and Neculai Stanciu,
Buzău, Romania.
(Vol. 51, No. 4, November 2013)
Prove that
n
(1)
(3)
n+3−Ln+2
e
≤
en+1−Fn+2 ≤
1X 1
n
Lk
1
n
k=1
n
X
k=1
1
Fk
!n
;
(2)
e
!n
;
(4)
en−Fn Fn+1
!n
n
1X 1
≤
;
n
L2
k=1 k
!n
n
1X 1
≤
.
n
Fk2
n+2−Ln Ln+1
k=1
Solution by Robinson Higuita, Medellı́n, Colombia.
First of all, we prove that if x ≥ 1, then ex ≤ ex , or equivalently, e1−x ≤ x1 . Let g(x) =
x
e − ex. Since g 0 (x) = ex − e > 0 for x > 1, we have that g is increasing in (1, ∞]. It is easy to
see that ex − ex ≥ 0 for x ≥ 1. This implies that ex ≤ ex all 1 ≤ x. Therefore, e1−x ≤ x1 . From
this and the inequality of arithmetic and geometric means, we have that for every sequence
{xk }1≤k≤n , with 1 ≤ xk , it holds that
!n
n
n
Pn
Y
X
1
1
1
en− k=1 xk = e1−x1 e1−x2 . . . e1−xn ≤
≤
,
xk
n
xk
k=1
k=1
namely,
n−
e
282
Pn
k=1 xk
n
≤
1X 1
n
xk
k=1
!n
.
(1)
VOLUME 53, NUMBER 3
ADVANCED PROBLEMS AND SOLUTIONS
On the other hand, it is known that (see for example pages 70, 77 and 78 in [1])
n
X
k=1
Lk = Ln+2 − 3,
n
X
k=1
n
X
Fk = Fn+2 − 1,
k=1
L2k = Ln Ln+1 − 2 and
Therefore, if in (1) we take xk = Lk , xk = Fk , xk =
en−(Ln+2 −3) = en−
n−(Fn+2 −1)
e
e
k=1
n−
=e
en−(Ln Ln+1 −2) = en−
n−Fn Fn+1
Pn
n−
=e
Lk
Pn
k=1 Fk
Pn
2
k=1 Lk
Pn
2
k=1 Fk
L2k
and xk =
Fk2 ,
n
≤
1X 1
n
Lk
!n
,
≤
1
n
!n
,
≤
n
1X 1
n
L2k
!n
,
≤
1
n
!n
,
respectively.
k=1
n
X
k=1
k=1
n
X
k=1
1
Fk
1
Fk2
n
X
Fk2 = Fn Fn+1 .
k=1
we obtain
References
[1] T. Koshy, Fibonacci and Lucas Numbers with Applications, John Wiley and Sons, Inc., New York, 2001.
Also solved by Kenneth B. Davenport, Dmitry Fleischman, Harris Kwong,
Hideyuki Ohtsuka, and the proposers.
On a Trigonometric Equation
H-745 Proposed by Kenneth B. Davenport, PA.
(Vol. 51, No. 4, November 2013)
√
Prove that (a2 − 1) cos(n + 3)θ − 2 a cos nθ = (a − 1)2 cos(n +
√1)θ, where a is the real
number satisfying a3 = a2 + a + 1 and θ is given by cos θ = (1 − a) a/2.
This problem was withdrawn in Vol. 52, No. 1, February 2014. Meanwhile,
Dmitry Fleischman, G. C. Greubel, Zbigniew Jakubczyk, Anastasios Kotronis,
and the proposer had provided solutions.
An Identity with Fibonomial Coefficients
H-746 Proposed by H. Ohtsuka, Saitama, Japan.
(Vol. 51, No. 4, November 2013)
n
Define the generalized Fibonomial coefficient
by
k F ;m
Fmn Fm(n−1) · · · Fm(n−k+1)
n
=
for 0 < k ≤ n
k F ;m
Fmk Fm(k−1) · · · Fm
AUGUST 2015
283
THE FIBONACCI QUARTERLY
n
n
= 1 and
= 0 (otherwise). Let εi = (−1)(m+1)i . For positive integers
with
k F ;m
0 F ;m
n, m and s prove that
X
i+j=2s
n
n
n
εi
= εs
.
i F ;m j F ;m
s F ;2m
Solution by the proposer.
n
Let
be the q-binomial coefficient. The q-binomial theorem is given by
k q
n
X
k k(k+1)/2
x q
k=0
n
Y
n
=
(1 + xq k ).
k q
(1)
k=1
Let x = αm(n+1) z, q = (β/α)m = (−α−2 )m . We have
n
X
k k(k+1)/2
x q
k=0
=
n
X
n
k
X
Y
n
1 − (β/α)m(n−r+1)
mk(n+1) k
−2 mk(k+1)/2
=
α
z (−α )
k q
1 − (β/α)mr
r=1
k=0
=
− β m(n−r+1)
· α2mr−m(n+1)
αmr − β mr
r=1
n
X
mk(k+1)/2 n
=
(−1)
zk ,
k F ;m
(−1)mk(k+1)/2 αmk(n+1)−mk(k+1) z k
k=0
n
X
(−1)mk(k+1)/2 z k
k
Y
Fm(n−r+1)
Fmr
r=1
k=0
k
Y
αm(n−r+1)
k=0
and
n
Y
(1 + xq k ) =
k=1
n
Y
(1 + αm(n+1) (β/α)mk z) =
k=1
n
Y
(1 + αm(n−k+1) β mk z).
k=1
Therefore, by (1), we obtain
n
n
X
Y
n
(−1)mk(k+1)/2
zk =
(1 + αm(n−k+1) β mk z).
k F ;m
k=0
(2)
k=1
Replacing z by −z in (2) we get
n
X
k=0
284
(−1)mk(k+1)/2+k
n
Y
n
zk =
(1 − αm(n−k+1) β mk z).
k F ;m
(3)
k=1
VOLUME 53, NUMBER 3
ADVANCED PROBLEMS AND SOLUTIONS
Using the identities (2) and (3), we have
n
n
X
Y
k n
2k
(−1)
z =
(1 − α2m(n−k+1) β 2mk z 2 )
k F ;2m
=
k=0
n
Y
k=1
(1 − αm(n−k+1) β mk z)(1 + αm(n−k+1) β mk z)
k=1

! n
n
X
X
n
n
=
(−1)mi(i+1)/2+i
z i  (−1)mj(j+1)/2
zj 
i F ;m
j F ;m
i=0
j=0


2n
X
X
n
n
2
2

 zr .
=
(−1)(m/2)(i +i+j +j)+i
i F ;m j F ;m
r=0
i+j=r
By comparing coefficients of z 2s we get
X
n
(m/2)(i2 +i+j 2 +j)+i n
s n
(−1)
= (−1)
.
i F ;m j F ;m
s F ;2m
i+j=2s
Here, since i + j = 2s, we have
m 2
m
(i + i + j 2 + j) = (i2 + i + (2s − i)2 + (2s − i)) = mi2 + 2ms2 − 2msi + ms.
2
2
Therefore, we have
X
n
mi+ms+i n
s n
(−1)
= (−1)
,
i F ;m j F ;m
s F ;2m
i+j=2s
which is the desired identity.
Solver’s note: From the identity in this problem, we obtain the following identity easily:
2
2n
X
2n
2n
εk
= εn
.
k F ;m
n F ;2m
k=0
Moreover, we obtain the following identity in the same manner:
X
n
n
n
n
s n
εa,a1
...
= (−1)
,
a F ;m a1 F ;m a2 F ;2m
ar F ;2r−1m
s F ;2r m
r
f (a,a1 ,...,ar )=2 s
where f (a, a1 , a2 , . . . , ar ) = a + a1 + 2a2 + · · · + 2r−1 ar and εi,j = (−1)m/2(i
2 +i+j 2 +j)+i
.
Also partially solved by Dmitry Fleischman.
AUGUST 2015
285