Solution of 12022. (Proposed by Mircea Merca), by Shalosh B. Ekhad, ShaloshBEkhad@gmail.com

Let F (n, k) be the summand. We have to prove that a(n) := ∑_{k=0}^{n−1} F (n, k) equals n. We cleverly come up with G(n, k) = − (1 − x^{n}) ⋯ (1 − x^{n−k+1}), that satisfies F (n + 1, k) − F (n, k) = G(n, k + 1) − G(n, k) (check!) . Hence

a(n+1)−a(n) = F (n+1, n)+∑_{k=0}^{n−1}( F (n+1, k)−F (n, k) ) = F (n+1, n)+∑_{k=0}^{n−1}( G(n, k+1)−G(n, k) )

= F (n + 1, n) + G(n, n) − G(n, 0) = (1 − x^{n}) ⋯ (1 − x) + (−(1 − x^{n}) ⋯ (1 − x)) − (−1) = 1 .

Since a(0) = 0, we have a(n) = n. The second identity is obtained from the first by replacing x by 1/x and simplifying.