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

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(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)−kn − k k − 1

3m X

j=1

(−2)−jn + 1 − 2k j − 1

 m − k 3m − j



= 0.

(Here, xk = Qk−1i=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)−kn − k k − 1



[x3m−1] (1 − x/2)n+1−2k(1 + x)m−k

= −1 2

X

r+s=n−1,s≤r

(−4)−1−sr s



[x3m−1] (1 − x/2)r−s(1 + x)m−s−1

=1 8

X

r+s=n−1,s≤r

(−1)s2−r−sr s



[x3m−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)sr s



[x3m−1] (2 − x)r−s(1 + x)m−s−1yr+s

= 2y[x3m−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[x3m−1](1 + x)m−1

X

r=0



(2 − x)y − y2 (1 + x)

r

= 2y[x3m−1] (1 + x)m−1 1 − (2 − x)y + y2/(1 + x)

= 2y[x3m−1] (1 + x)m

(1 + x) − (2 − x)(1 + x)y + y2

= 2y

(1 − y)2[x3m−1] (1 + x)m 1 + ux + (u2− u)x2

(2)

where we define u = 1/(1 − y). Now 1

1 + ux + (u2− u)x2 = 1

(1 − αx)(1 − βx) =

X

k=0

akxk

where

ak= αk+1− βk+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

 α2m+j− β2m+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).

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