In previous pages we have looked at various ways to sum the powers of
the natural numbers: powers of 1
and 2. Here we
will generalise and find one (of many) formulae to give us the sums of
powers with much less work.

Some of the techniques we examined, worked only for some of the powers.
The technique
of summation works for all powers.

In general, the sum of the (k+1) terms is the sum of the k terms
plus the (k+1) term:
[1.1]

Expanding the left-hand side, using the Binomial Theorem, we get:
[1.2]

Replacing the left-hand side of 1.1 with the right-hand side of 1.2, we
get:

[1,3]

As expected the k^{m+1} terms will cancel.

Leaving the k^{m} sum where it is, we can move the rest to the right-hand
side;
[1.4]

More generally, we can say: [1.5]
This
is a recursive formula, of course, so you can find the sum of the m
powers only when you know the sum of the (m-1) powers. By setting m=0,
1, 2, 3...m, you can find the sum of any m power. .

Let us use the formula, by setting m=0.

Because the lower sum, r=2 exceeds the upper 1, then all that remains
is (n+1) which is the sum of 1's from 0 to n. ∑^{n}_{0}1=(n+1)!

Let m=3, to find the sum of the first n cubes. Substituting in the
equation:

We find:

Giving us the formula:

The
sum of the cubes of the first n natural numbers is the square of the
formula for the first n natural numbers, so it is easy to remember!

Examples
of Powers of the Natural Numbers

The following table collects information about the coefficients of the sums of the first n powers of the natural numbers.

Ken's book is packed with examples and explanations that enable you to discover more than 150 techniques to speed up your arithmetic and increase your understanding of numbers. Paperback and Kindle: