Sum of Powers

The sum S_n(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) – Σi=1n–1Si(p–1)
=	nSn(p–1) – Σi=1nSi(p–1) + Sn(p–1)
=	(n+1)Sn(p–1) – Σi=1nSi(p–1)

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

Sn(1) =	(n+1)Sn(0) – Σi=1nSi(0)
=	n(n+1) –Σi=1ni
=	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, S_n(2)

Sn(2) =	(n+1)Sn(1) – Σi=1nSi(1)
=	n(n+1)2/2 – (Σi=1ni2 + Σi=1ni)/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, S_n(3)

Sn(3) =	(n+1)Sn(2) –Σi=1nSi(2)
=	n(n+1)2(2n+1)/6 – (2Σi=1ni3 + 3Σi=1ni2 + Σi=1ni)/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

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