Critique on a proof by induction that $\sum\limits_{i=1}^n i^2= n(n+1)(2n+1)/6$? I need to make the proof for this
1:$$1^2 + 2^2 + 3^2 + ... + n^2=\frac{(n(n+1)(2n+1))}{6}$$
By mathematical induction I know that,
If P(n) is true for $n>3^2$ then P(k) is also true for k=N and also P(k+1) must be true. 
Then,
2:$$1^2 + 2^2 + 3^2 + ... + k^2=\frac{(k(k+1)(2k+1))}{6}$$
Substitution:
3:$$1^2 + 2^2 + 3^2 + ... +n^2+ (n+1)^2=\frac{((n+1)((n+1)+1)(2(n+1)+1))}{6}$$
Questions
a: Do anyone see notation errors in this post?
b:  It is right so far?

For the third step, notice that
$$ 1^2 + 2^2 + 3^2 + ... +n^2+ (n+1)^2 = \frac{(n(n+1)(2n+1))}{6}+(n+1)^2 $$
$$ =  \frac{(n(n+1)(2n+1))}{6}+\frac{6(n+1)^2}{6}$$
$$ = \frac{(n+1)(n(2n+1)+6(n+1))}{6}=\dots $$
Can you finish it?
