How does this summation equal \frac{n(n+1)(2n+1)}{6}?

[vanmaiden]

Homework Statement

I'm not very proficient with LAtex, so I'll try to translate this mess the best I can. It's a summation
\sum i² (on the bottom, there would be an "i = 1") (on the top, there would be an "n") this summation equals \frac{n(n+1)(2n+1)}{6}

In the summation, basically, "i" starts off at one and goes to "n"

Why does this summation equal \frac{n(n+1)(2n+1)}{6}?

Homework Equations

The definition of a definite integral? I found this summation through doing definite integrals using Riemann sums. I found the answer to the summation online, but I wanted to know how one arrives at the answer.

The Attempt at a Solution

So far, I have done this:

S = 1² + 2² + 3² + ... + (n - 1)² + n²
S = n² + (n - 1)² + (n - 2)² + ... + 2² + 1²

I just wrote out a part of the summation, and under it, I did the same summation part but in reverse.

Then...I added the two summations to get

2S = n² + 1 + (n - 1)² + 4 + (n - 2)² + 9 + ... + (n - 1)² + 4 + n² + 1

I saw somewhat of a pattern, but it was hard to explain and I had to stop here. If someone could give me a link to an explanation, or if someone could fit an explanation in a response or two as to why this summation equals \frac{n(n+1)(2n+1)}{6}, that would be fantastic. Thank you in advance!

 

[eumyang]

Try this: http://www.trans4mind.com/personal_development/mathematics/series/sumNaturalSquares.htm"