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)ik i kn− in k+ 1

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)ⁱk 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

[t^k+1]((1 + t)ⁿ− 1)

k =

k

X

i=0

k i



(−1)ⁱ[t^k+1](1 + t)^n(k−i)=

k

X

i=0

(−1)ⁱk i

kn− in k+ 1

 .

On the other hand

[t^k+1]((1 + t)ⁿ− 1)^k= [t^k+1]



t

n

X

j=1

n j

 t^j−1





k

= [t]





n

X

j=1

n j

 t^j−1





k

= [t]

 n+n

2

 t+ o(t)

k

= n^k[t]



1 + n − 1 2 t+ o(t)

k

= kn^k(n − 1)

2 .

Therefore

k

X

i=0

(−1)ⁱk i

kn− in k+ 1



= kn^k(n − 1)

2 .