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Proof that $\sum\limits_{k=1}^nk^2 = \frac{n(n+1)(2n+1)}{6}$?
Finite Sum of Power?

I know that the sum of the squares of the first n natural numbers is $\frac{n(n + 1)(2n + 1)}{6}$. I know how to prove it inductively. But how, presuming I have no idea about this formula, should I determine it? The sequence $a(n)=1^2+2^2+...+n^2$ 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?

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marked as duplicate by J. M., Pedro Tamaroff, MJD, Martin Sleziak, Ross Millikan Aug 16 '12 at 18:17

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

The first chapter of Concrete Mathematics by Graham, Knuth, and Patashnik presents about seven different techniques for deriving this identity, so you might be interested to look at that. – MJD Aug 16 '12 at 18:19
Coincidentally, I uploaded a web page recently regarding this topic. There I don't prove the formula (or even present it), but I provide the student a geometric strategy for finding it. – John Joy Aug 19 '14 at 16:18

2 Answers 2

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

Consider the following sum: $$\sum_{i=1}^n((1+i)^3-i^3).$$

First, looking at it as a telescoping sum, you will get $$\sum_{i=1}^n((1+i)^3-i^3)=(1+n)^3-1.$$

On the other hand, you also have $$\sum_{i=1}^n((1+i)^3-i^3)=\sum_{i=1}^n(3i^2+3i+1)=3\sum_{i=1}^ni^2+3\sum_{i=1}^ni+n.$$

Using these two expressions, and the fact that $\sum_{i=1}^ni=\frac{n(n+1)}{2}$, you can now solve for $\sum_{i=1}^ni^2$.

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Shouldn't it be $3\sum_{i=1}^ni^2+3\sum_{i=1}^ni+\sum_{i=1}^n1$ instead of $3\sum_{i=1}^ni^2+3\sum_{i=1}^ni+1$? – George Apriashvili Oct 15 '14 at 11:54

If you know it is a cubic, you can just guess that it is $an^3+bn^2+cn+d$. You might guess this either by analogy with the sum of first powers being a square or analogy with integration. Then you can calculate the first four terms and solve for $a,b,c,d$. Another way is to say $a(n+1)^3+b(n+1)^2+c(n+1)+d-(an^3+bn^2+cn+d)=(n+1)^2$ and equate like powers of $n$.

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