# Sums of Squares

In the past, I wrote about how the sum of the first N cubes is the square of the sum of the first N numbers. In it I remarked how finding the sum of the first N numbers raised to some power is easy to do using induction and telescoping series. This is fine, and probably the easiest way of doing it, but it seems a little indirect. So today I thought I’d try and come up with a more direct way of doing it, and here it is.

Define Q to be the sum of the first N squares:

$Q = 1^2 + 2^2 + \ldots + N^2$

The key is that it can be rewritten like this:

$Q = 1 + (2+2) + (3+3+3) + \ldots + (\underbrace{N+\ldots+N}_{N \mbox{ times}})$

Rearranging the terms of the sum, we have:

$Q = (1+2+3+\ldots+N) + (2+3+\ldots+N) + (3+4+\ldots+N) +\ldots + N$

And so:

$Q = N(1+2+\ldots+N) - [ 1 + (1+2) + (1+2+3) + \ldots + (1+2+\ldots+(N-1))]$

Using the equation for sum of the first N numbers, we have:

$Q = \frac{1}{2}N^2(N+1) - \sum_{i=1}^{N-1}\frac{1}{2}i(i+1)$

Expanding out the sum:

$Q = \frac{1}{2}N^2(N+1) - \frac{1}{2}\left[\sum_{i=1}^{N-1}i^2 + i\right]$

Expressing the sum of squares in terms of Q again, and applying the formula for the first N-1 numbers, we have:

$Q = \frac{1}{2}N^2(N+1) - \frac{1}{2}\left[Q-N^2 + \frac{1}{2}N(N-1)\right]$

Putting the equation in terms of Q yields:

$\frac{3}{2}Q = \frac{1}{4}N\left[2N(N+1) + 2N - (N-1)\right]$

Thus,

$Q = \frac{1}{6}N(N+1)(2N+1)$