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# We have that [tk+1]((1 + t)n− 1) k = k X i=0 k i  (−1)i[tk+1](1 + t)n(k−i)= k X i=0 (−1)ik i kn− in k+ 1

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Problem 11862

(American Mathematical Monthly, Vol.122, October 2015)

Proposed by David A. Cox and Uyen Thieu (USA).

For positive integers n and k, evaluate

k

X

i=0

(−1)ik i

kn− in k+ 1

 .

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

We have that

[tk+1]((1 + t)n− 1)

k =

k

X

i=0

k i



(−1)i[tk+1](1 + t)n(k−i)=

k

X

i=0

(−1)ik i

kn− in k+ 1

 .

On the other hand

[tk+1]((1 + t)n− 1)k= [tk+1]

t

n

X

j=1

n j

 tj−1

k

= [t]

n

X

j=1

n j

 tj−1

k

= [t]

 n+n

2

 t+ o(t)

k

= nk[t]



1 + n − 1 2 t+ o(t)

k

= knk(n − 1)

2 .

Therefore

k

X

i=0

(−1)ik i

kn− in k+ 1



= knk(n − 1)

2 .



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