[lupire]

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

Number of necklaces of partitions of n+1 labeled beads.

[vismit2000]

Mathologer video on the same: https://www.youtube.com/watch?v=4AuV93LOPcE