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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.

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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
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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. mathuprising.comlu.com/sumofsquares –  John Joy yesterday

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 $$\sum_{i=1}^n((1+i)^3-i^3)=\sum_{i=1}^n(3i^2+3i+1)=3\sum_{i=1}^ni^2+3\sum_{i=1}^ni+1.$$

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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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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