How to get to the formula for the sum of squares of first n numbers? [duplicate]

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I know that the sum of the squares of the first n natural numbers is n(n+1)(2n+1)6. I know how to prove it inductively. But how, presuming I have no idea about this formula, should I determine it? The sequence a(n)=12+22+...+n2 is neither geometric nor arithmetic. The difference between the consecutive terms is 4, 9, 16 and so on, which doesn’t help. Could someone please help me and explain how should I get to the well known formula assuming I didn’t know it and was on some desert island?

Answer

This is proven, for example, in Stewart’s Calculus:

Consider the following sum:
ni=1((1+i)3i3).

First, looking at it as a telescoping sum, you will get
ni=1((1+i)3i3)=(1+n)31.

On the other hand, you also have
ni=1((1+i)3i3)=ni=1(3i2+3i+1)=3ni=1i2+3ni=1i+n.

Using these two expressions, and the fact that ni=1i=n(n+1)2, you can now solve for ni=1i2.

Attribution
Source : Link , Question Author : Straightfw , Answer Author : M Turgeon

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