(1)Problem 12032 (American Mathematical Monthly, Vol.125, March 2018) Proposed by D

Problem 12032

(American Mathematical Monthly, Vol.125, March 2018) Proposed by D. Galante and A. Plaza (Spain).

For a positive integer n, compute

n

X

j=0 n

X

k=j

(−1)^k−j k 2j

n k

 2^n−k.

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

Solution. We have that

n

X

j=0 n

X

k=j

(−1)^k−j k 2j

n k

 2^n−k=

n

X

k=0

(−1)^kn k

 2^n−k

⌊k/2⌋

X

j=0

(−1)^j k 2j



=

n

X

k=0

(−1)^kn k



2^n−kRe (1 + i)^k

= Re

n

X

k=0

n k



2^n−k(−1 − i)^k

!

= Re ((2 − 1 − i)ⁿ) = Re ((1 − i)ⁿ)

= 2^n/2Re e^−inπ4 

= 2^n/2cosnπ 4



which is always an integer number (see sequence A146559 in OEIS).