Show that for all positive integers m and n, n X k=1 (−4)−kn − k k − 1 3m X j=1 (−2)−jn + 1 − 2k j − 1

Problem 11914

(American Mathematical Monthly, Vol.123, May 2016) Proposed by R. Chapman (UK) and R. Tauraso (Italy).

Show that for all positive integers m and n,

n

X

k=1

(−4)^−kn − k k − 1

^3m X

j=1

(−2)^−jn + 1 − 2k j − 1

 m − k 3m − j



= 0.

(Here, ^x_k
= Q^k−1_i=0(x − i)/k! for x ∈ R).

Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.

Let S(m, n) be the left-hand side then S(m, n) = −1

2

n

X

k=1

(−4)^−kn − k k − 1



[x^3m−1] (1 − x/2)^n+1−2k(1 + x)^m−k

= −1 2

X

r+s=n−1,s≤r

(−4)^−1−sr s



[x^3m−1] (1 − x/2)^r−s(1 + x)^m−s−1

=1 8

X

r+s=n−1,s≤r

(−1)^s2^−r−sr s



[x^3m−1] (2 − x)^r−s(1 + x)^m−s−1
.

For each positive integer m, introduce the generating function Fm(y) = 8

∞

X

n=1

S(m, n)(2y)ⁿ.

Thus

Fm(y) = 2y

∞

X

r=0 r

X

s=0

(−1)^sr s



[x^3m−1] (2 − x)^r−s(1 + x)^m−s−1y^r+s

= 2y[x^3m−1](1 + x)^m−1

∞

X

r=0

(y(2 − x))^r

r

X

s=0

r s

  −y

(1 + x)(2 − x)

s

= 2y[x^3m−1](1 + x)^m−1

∞

X

r=0



(2 − x)y − y² (1 + x)

^r

= 2y[x^3m−1] (1 + x)^m−1 1 − (2 − x)y + y²/(1 + x)

= 2y[x^3m−1] (1 + x)^m

(1 + x) − (2 − x)(1 + x)y + y²

= 2y

(1 − y)²[x^3m−1] (1 + x)^m 1 + ux + (u²− u)x²

where we define u = 1/(1 − y). Now 1

1 + ux + (u²− u)x² = 1

(1 − αx)(1 − βx) =

∞

X

k=0

akx^k

where

ak= α^k+1− β^k+1

α − β with α = −u +√

4u − 3u²

2 , β = −u −√

4u − 3u²

2 .

In order to prove that Fm(y) = 0, and therefore S(m, n) = 0, we need to show that for each positive integer m,

0 =

m

X

j=0

m j



a3m−1−(m−j)=

m

X

j=0

m j

 α^2m+j− β^2m+j

α − β = (α²(α + 1))^m− (β²(β + 1))^m α − β

which holds because it is straightforward to verify that α²(α + 1) = u(α + 1)(β + 1) = β²(β + 1).