How do I prove by induction? For example if i wanted to prove: 
$1^2 + \dots + n^2 = \frac {n(n + 1)(2n + 1)} {6}$
by induction.
I'm not sure where to start. 
Thanks. 

Prove for the base case, n=1:
$$1^{2} = \frac{1(1+1)(2+1)}{6} = \frac{2\cdot3}{6}=1$$
The "sum" of just $1^2$ is indeed 1. Base case proven.
Now for the induction step, proving that it holds for n+1:
Observe that for $1^{2} + 2^{2} +... + n^{2}$, the sum including $n+1$ equals the original summation plus the $n+1$ term: $1^{2} + 2^{2} + ... + n^{2} + (n+1)^{2}$, i.e. equals $ \frac{n(n+1)(2n+1)}{6} + (n+1)^{2}$.
So all we need to do is show that $\frac{n(n+1)(2n+1)}{6} + (n+1)^{2} = \frac{(n+1)((n+1)+1)(2(n+1)+1)}{6}$.
(Edit: To be clear: $\frac{(n+1)((n+1)+1)(2(n+1)+1)}{6}$ is $\frac{n(n+1)(2n+1)}{6}$ with $n+1$ substituted in for $n$.)
Let's do that now:
$$\frac{n(n+1)(2n+1)}{6} + (n+1)^{2} = \frac{n(n+1)(2n+1)}{6} + \frac{6(n+1)^{2}}{6}$$
$$ = \frac{n(n+1)(2n+1)+6(n+1)^2}{6}$$
$$ = \frac{n(n+1)(2n+1)+6(n^{2}+2n+1)}{6}$$
$$ = \frac{n(2n^{2}+3n+1)+6n^{2}+12n+6)}{6}$$
$$ = \frac{2n^{3}+3n^{2}+n+6n^{2}+12n+6}{6}$$
$$ = \frac{2n^{3}+9n^{2}+13n+6}{6}$$
Having simplified this, we'll now show that it is equal to $\frac{(n+1)((n+1)+1)(2(n+1)+1)}{6}$:
$$\frac{(n+1)((n+1)+1)(2(n+1)+1)}{6} = \frac{(n+1)(n+2)(2n+3)}{6}$$
$$ = \frac{(n^{2}+3n+2)(2n+3)}{6}$$
$$ = \frac{2n^{3}+6n^{2}+4n+3n^{2}+9n+6}{6}$$
$$ = \frac{2n^{3}+9n^{2}+13n+6}{6}$$
And we're done!
