(1)Problem 11919 (American Mathematical Monthly, Vol.123, June-July 2016) Proposed by A

Problem 11919

(American Mathematical Monthly, Vol.123, June-July 2016) Proposed by A. Alt (USA).

For positive integers m, n and k with k≥ 2, prove

n

X

i₁=1

· · ·

n

X

ik=1

(min {i1, . . . , ik})^m=

m

X

i=1

(−1)^m−im i



(n + 1)ⁱ− nⁱ

n

X

j=1

j^k+m−i.

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

Solution. In the set {1, 2, . . . , n}^k there are ^k_i
(n − j)^k−i points whose minimal coordinate value j appears i times. Hence

n

X

i₁=1

· · ·

n

X

ik=1

(min {i1, . . . , ik})^m=

n

X

j=1

j^m

k

X

i=1

k i



(n − j)^k−i.

On the other hand,

m

X

i=1

(−1)^m−im i



(n + 1)ⁱ− nⁱ

n

X

j=1

j^k+m−i=

n

X

j=1

j^k

m

X

i=0

m i



(n + 1)ⁱ− nⁱ
(−j)^m−i

=

n

X

j=1

j^k((n + 1 − j)^m− (n − j)^m)

=

n

X

j=1

j^k(n − (j − 1))^m−

n−1

X

j=0

j^k(n − j)^m

=

n−1

X

j=0

(j + 1)^k(n − j)^m−

n−1

X

j=0

j^k(n − j)^m

=

n−1

X

j=0

(n − j)^m (j + 1)^k− j^k

=

n−1

X

j=0

(n − j)^m

k

X

i=1

k i

 j^k−i

=

n

X

j=1

j^m

k

X

i=1

k i



(n − j)^k−i.