- #1
twoflower
- 368
- 0
Hi,
I wonder how could I get to this formula:
[tex]
1^2 + 2^2 + 3^2 + ... + n^2 = \frac{n(n+1)(2n+1)}{6}
[/tex]
I'd like to know it, because I suspect I will keep this formula in head and because it is quite useful I'd like to be able to write it from scratch.
Thank you.
I wonder how could I get to this formula:
[tex]
1^2 + 2^2 + 3^2 + ... + n^2 = \frac{n(n+1)(2n+1)}{6}
[/tex]
I'd like to know it, because I suspect I will keep this formula in head and because it is quite useful I'd like to be able to write it from scratch.
Thank you.