Sum of Powers

The sum Sn(p) of the series of n integers raised to an arbitrary power p may be expressed in terms of sums involving the next lower power p-1:

Sn(p) = 1p + 2p + 3p + ... + (n-1)p + np = 1p-1 + 2p-1 + 3p-1 + ... + (n-1)p-1 + np-1 + 2p-1 + 3p-1 + ... + (n-1)p-1 + np-1 + 3p-1 + ... + (n-1)p-1 + np-1 . . . . + (n-1)p-1 + np-1 + np-1

This equals n times the sum of the series of integers raised to the next lower power minus the sums of the smaller series omitted at the beginning of each row:

Sn(p) = nSn(p-1) - n-1Σ1Si(p-1) = nSn(p-1) - n Σ1Si(p-1) + Sn(p-1) = (n+1)Sn(p-1) - n Σ1Si(p-1)

To find the sum of integers, Sn(1), note that Sn(0) = n

Sn(1) = (n+1)Sn(0) - n Σ1Si(0) = n(n+1) - n Σ1i = n(n+1) - Sn(1) 2Sn(1) = n(n+1)

 Sn(1) = n(n+1) 2

This result may be used to find the sum of squares, Sn(2)

Sn(2) = (n+1)Sn(1) - n Σ1Si(1) = n(n+1)2/2 - ( n Σ1i2 + n Σ1i)/2 2Sn(2) = n(n+1)2 - Sn(2) - n(n+1)/2 3Sn(2) = n(n+1)[(n+1) - 1/2]

 Sn(2) = n(n+1)(2n+1) 6

This result may be used to find the sum of cubes, Sn(3)

Sn(3) = (n+1)Sn(2) - n Σ1Si(2) = n(n+1)2(2n+1)/6 - (2 n Σ1i3 + 3 n Σ1i2 + n Σ1i)/6 6Sn(3) = n(n+1)2(2n+1) - 2Sn(3) - n(n+1)(2n+1)/2 - n(n+1)/2 8Sn(3) = n(n+1)[(n+1)(2n+1) - (2n+1)/2 - 1/2]     = n(n+1)[(n+1)(2n+1) - (n+1)]     = n(n+1)2(2n)

 Sn(3) = n2(n+1)2 4