Here's the question you clicked on:

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

• This Question is Closed
1. across Group Title

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

2. helder_edwin Group Title

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

3. across Group Title

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

4. waterineyes Group Title

Can I provide the link or not here ??

5. waterineyes Group Title

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

6. experimentX Group Title

i think this was done in physics section ...

7. waterineyes Group Title

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

8. helder_edwin Group Title

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

9. experimentX Group Title

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

10. across Group Title

Links are more appreciated than derivations here. ^^

11. waterineyes Group Title

http://www.ilovemaths.com/3sequence.asp May be it will help you.

12. waterineyes Group Title

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

13. experimentX Group Title
14. experimentX Group Title

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

15. asnaseer Group Title

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 Group Title

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