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Prove that for all $n \in \mathbb N$, $0^2 + 1^2 + 2^2 + \ldots + n^2 = \frac {n(n + 1)(2n + 1)}{6}$.

Define $ p(n)=0^2 + 1^2 + 2^2 + \ldots + n^2$. Then: \begin{align*}p(n + 1)&=0^2 + 1^2 + 2^2 + \ldots + n^2 + (n + 1)^2\\ &= \frac {n(n + 1)(2n + 1) + 6(n + 1)^2}{6}\\ &= \dots\\ &= \frac {(n + 1) (n + 2)(2n + 3)}{6} \end{align*}

How do I quickly manipulate $\displaystyle \frac {n(n + 1)(2n + 1) + 6(n + 1)^2}{6}$ to end up with $ \displaystyle \frac {(n + 1) (n + 2)(2n + 3)}{6}$? Any tricks or tips for something like that?

Thanks

Alex Schiff
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New Orleans
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  • Take $(n+1)$ common from the two terms, multiply, and then again factorise the resulting quadratic by the trinomial technique. Have you learnt how to factorise a quadratic trinomial?If not, a google search on factorising trinomials would be a nice thing to do – Shubham May 22 '14 at 19:25
  • What are some general factoring techniques that are oft used in such situations? Or what are th things that I can look out for in general? Getting from one polynomial form to another is such a pain. – New Orleans May 22 '14 at 19:28
  • You have the terms you want to manipulate and the term you want to end up with. You want to preserve the factors in the term you end up with from what you're working with and use everything else to introduce the rest of the factors. – Pockets May 22 '14 at 19:30
  • The trinomial technique is a nice one. For fast working, you've got to learn a lot of stuff, e.g. start with expansions. If you want to get deeper knowledge, go for the remainder and factor theorems for polynomials(helpful for factorising polynomials of degree 3 or larger) – Shubham May 22 '14 at 19:33

1 Answers1

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how about taking common $n+1$ and then adding $$n(2n+1) + 6(n+1) = 2n^2 + 4n + 3n + 6 = (n+2)(2n + 3)$$

S L
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