This is a small proof I wrote up working through the second chapter of Michael Spivak's Calculus. I'm fairly certain that it is correct but it is my first attempt at writing out a "well-formatted" proof and thus I am worried if there are any problems with layout or style.
We are asked to prove the following statement using mathematical induction,
$$1^2+...+n^2 = \frac{n(n+1)(2n+1)}{6}.$$
A quick check convinces us that it holds true for $n=1$ and, as follows, we assume the expression to be true for some $n=k$. Now, we wish to show that
$$1^2+...+k^2+(k+1)^2 = \frac{(k+1)(k+2)(2k+3)}{6},$$
which we do by using our just stated assumption as follows
$$[1^2+...+k^2]+(k+1)^2 = [\frac{k(k+1)(2k+1)}{6}] + (k+1)^2.$$
Then, on the right hand side, we factor out $(k+1)$ and simplify, giving us the expression
$$(k+1)\frac{2k^2+7k+6}{6}.$$
Using, for example, polynomial long division, we can factor out $k+2$ from $2k^2+7k+3$, leaving us with $(k+2)(2k+3)$ and finally $$\frac{(k+1)(k+2)(2k+3)}{6}. $$ $$\tag*{$\blacksquare$}$$