Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Possible Duplicate:
How do I come up with a function to count a pyramid of apples?
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?

share|cite|improve this question

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
You can find a formal approach for even higher powers here as well as a nice geometric approach here. – Jan Mar 10 at 12:08

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 $\sum_{i=1}^ni^2$.

share|cite|improve this answer
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 $n$.

share|cite|improve this answer

Not the answer you're looking for? Browse other questions tagged or ask your own question.