Problem 12160
(American Mathematical Monthly, Vol.127, February 2020) Proposed by H. Ohtsuka (Japan) and R. Tauraso (Italy).
LetFn be thenth Fibonacci number, and let Ln be thenth Lucas number. Prove
n
X
k=0
2n + 1 n − k
F2k+1= 5n and
n
X
k=0
2n + 1 n − k
L2k+1=
n
X
k=0
2k k
5n−k
for alln ∈ N.
Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
Solution. Let
f (x, y) :=
∞
X
n=0
xn
n
X
k=0
2n + 1 n − k
y2k+1=
∞
X
k=0
y2k+1
∞
X
n=k
2n + 1 n − k
xn
= y
∞
X
k=0
(xy2)k
∞
X
n=0
2n + (2k + 1) n
xn
= 2y
√1 − 4x(1 +√ 1 − 4x)
∞
X
k=0
4xy2 (1 +√
1 − 4x)2
k
= 2y
√1 − 4x(1 +√
1 − 4x)
1 − (1+√4xy1−4x)2 2
= y(1 +√
1 − 4x)
√1 − 4x 1 +√
1 − 4x − 2x(1 + y2) where we used the known identity
X∞ n=0
2n + m n
xn= 1
√1 − 4x
2
1 +√ 1 − 4x
m
(see (5.72) in Concrete Mathematics by D. E. Knuth, O. Patashnik, and R. Graham).
Since
F2k+1=a2k+1− b2k+1
√5 and L2k+1= a2k+1+ b2k+1 with a := (1 +√
5)/2 and b := (1 −√
5)/2, it follows that X∞
n=0
xn
n
X
k=0
2n + 1 n − k
F2k+1= f (x, a) − f(x, b)
√5 ,
and ∞
X
n=0
xn
n
X
k=0
2n + 1 n − k
L2k+1= f (x, a) + f (x, b).
On the other hand
g(x) :=
X∞ n=0
5nxn = 1 1 − 5x. and
h(x) :=
X∞ n=0
xn
n
X
k=0
2k k
5n−k= 1
√1 − 4x (1 − 5x). Hence, in order to show our identities, it remains to check that
f (x, a) − f(x, b) =√
5 g(x) and f (x, a) + f (x, b) = h(x)
which can be easily verified.
