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