## across 3 years ago Let's go over the derivation of$\sum_{i=1}^{n}i^2=\frac{n(n+1)(2n+1)}{6}$

1. across

This could follow from Gauss' assertion that$\sum_{i=1}^{n}i=\frac{n(n+1)}{2}$

2. helder_edwin

i don't remember the details. but i'm pretty sure u can find this in Spivaks' Calculus

3. across

It's easy to prove this by induction with your eyes closed, but I'm wondering how it's derived.

4. waterineyes

Can I provide the link or not here ??

5. waterineyes

@across can I provide here the link or I have to derive here full??

6. experimentX

i think this was done in physics section ...

7. waterineyes

This is nothing but sum of squares of first n natural numbers..

8. helder_edwin

yes i know the proof is easy. i suggested spivak's book because i remember seeing the derivation there

9. experimentX

i would prefer geometrical visualization instead ... i remember seeing one is mit ocw single variable calculus. the volume of pyramid ...

10. across

Links are more appreciated than derivations here. ^^

11. waterineyes

12. waterineyes

http://www.9math.com/book/sum-squares-first-n-natural-numbers Or you can check this also..

13. experimentX
14. experimentX

difference of power is very power technique ... it can be used to find the sums of n^3 and n^4 also ....

15. asnaseer

There is a very beautiful visual proof of this that @Ishaan94 came across when he asked this question: http://openstudy.com/updates/5002f7dae4b0848ddd66eea4

16. experimentX

very interesting proof indeed !! $3 \left ( \sum_{n=1}^\infty n^2 \right ) = (1 + 2 +3 +... + n)(2n+1)$