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

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across
 4 years ago
Best ResponseYou've already chosen the best response.0This could follow from Gauss' assertion that\[\sum_{i=1}^{n}i=\frac{n(n+1)}{2}\]

helder_edwin
 4 years ago
Best ResponseYou've already chosen the best response.0i don't remember the details. but i'm pretty sure u can find this in Spivaks' Calculus

across
 4 years ago
Best ResponseYou've already chosen the best response.0It's easy to prove this by induction with your eyes closed, but I'm wondering how it's derived.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Can I provide the link or not here ??

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@across can I provide here the link or I have to derive here full??

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.1i think this was done in physics section ...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0This is nothing but sum of squares of first n natural numbers..

helder_edwin
 4 years ago
Best ResponseYou've already chosen the best response.0yes i know the proof is easy. i suggested spivak's book because i remember seeing the derivation there

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.1i would prefer geometrical visualization instead ... i remember seeing one is mit ocw single variable calculus. the volume of pyramid ...

across
 4 years ago
Best ResponseYou've already chosen the best response.0Links are more appreciated than derivations here. ^^

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0http://www.ilovemaths.com/3sequence.asp May be it will help you.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0http://www.9math.com/book/sumsquaresfirstnnaturalnumbers Or you can check this also..

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.1http://www.trans4mind.com/personal_development/mathematics/series/sumNaturalSquares.htm#general_term

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.1difference of power is very power technique ... it can be used to find the sums of n^3 and n^4 also ....

asnaseer
 4 years ago
Best ResponseYou've already chosen the best response.1There is a very beautiful visual proof of this that @Ishaan94 came across when he asked this question: http://openstudy.com/updates/5002f7dae4b0848ddd66eea4

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.1very interesting proof indeed !! \[ 3 \left ( \sum_{n=1}^\infty n^2 \right ) = (1 + 2 +3 +... + n)(2n+1)\]
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