Question

Let's go over the derivation of\[\sum_{i=1}^{n}i^2=\frac{n(n+1)(2n+1)}{6}\]

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

Can I provide the link or not here ??

Answer 5

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

Answer 6

i think this was done in physics section ...

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

Links are more appreciated than derivations here. ^^

Answer 11

http://www.ilovemaths.com/3sequence.asp

May be it will help you.

Answer 12

http://www.9math.com/book/sum-squares-first-n-natural-numbers

Or you can check this also..

Answer 13

http://www.trans4mind.com/personal_development/mathematics/series/sumNaturalSquares.htm#general_term

Answer 14

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

Answer 15

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

Answer 16

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