Problem 11767
(American Mathematical Monthly, Vol.121, March 2014)
Proposed by M. Merca (Romania).
Prove that
X(1 + t1+ t2+ · · · + tn)!
(1 + t1)!t2! · · · tn! = 2n− Fn
where the sum is over all nonnegative integer solutions to t1+ 2t2+ · · · + ntn= n and Fk is the kth Fibonacci number.
Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
Let F (x, y) = (exy− 1)e1−xx2 y for x, y ∈ (−1, 1) then
F (x, y) = (exy− 1)
∞
Y
k=2
ex2y =
∞
X
t1=0
(xy)1+t1 (1 + t1)!
∞
Y
k=2
∞
X
tk=0
(xky)tk tk! =
∞
X
n=0
X
∗
y1+t1+t2+···+tn (1 + t1)!t2! · · · tn!xn+1
whereP
∗ is over all nonnegative integer solutions to t1+ 2t2+ · · · + ntn= n. Hence
∞
X
n=0
X
∗
(1 + t1+ t2+ · · · + tn)!
(1 + t1)!t2! · · · tn! xn+1=
∞
X
m=0
∂m(F (x, y))
∂ym y=0
=
∞
X
m=0 m
X
k=0
m k
∂m−k(exy− 1)
∂ym−k y=0
· ∂k(e1−xx2 y)
∂yk y=0
=
∞
X
m=0 m
X
k=0
m k
(xm−k− δm,k) · x2k (1 − x)k
=
∞
X
m=0
xm
m
X
k=0
m k
xk
(1 − x)k − x2m (1 − x)m
!
=
∞
X
m=0
xm
1 + x
1 − x
m
− x2m (1 − x)m
=
∞
X
m=0
xm− x2m (1 − x)m
=
∞
X
m=1
x 1 − x
m
−
∞
X
m=1
x2 1 − x
m
= x
1 − 2x− x2 1 − x − x2
=
∞
X
n=0
(2n− Fn) xn+1,
which implies that for any nonnegative integer n,
X
∗
(1 + t1+ t2+ · · · + tn)!
(1 + t1)!t2! · · · tn! = 2n− Fn.
