And since it's discrete, you can use finite differences. a = sum_1 n^3 / 2^n = sum_0 (n+1)^3 / 2^(n+1) = (1/2) (1 + sum_1 (3n^2 + 3n + 1)/2^n)

And since it's discrete, you can use finite differences. a = sum_1 n^3 / 2^n = sum_0 (n+1)^3 / 2^(n+1) = (1/2) (1 + sum_1 (3n^2 + 3n + 1)/2^n) b = sum_1 n^2 / 2^n = (1/2) (1+ b + sum_1 (2n +1)/2^n) c = sum_1 n / 2^n = (1/2) (1+ c + sum_1 1 / 2^n) d = sum_1 1 / 2^n = sum_0 1 / 2^(n+1) = (1/2) (1 + d) = 1 d = 1 = 1 c = d + 1 = 1+1 = 2 b = d + 2c + 1 = 1+1+4 = 6 a = d + 3c + 3b + 1 = 1+1+6+18 = 26 In general, f_k = sum n^k/2^n = (*k*th row of Pascal's triangle)•(f_0, ..., f_{k-1},1) https://oeis.org/A000629 https://oeis.org/A000629 Number of necklaces of partitions of n+1 labeled beads.