Induction proof for a sum of squares

Use mathematical induction to prove that

$\sum_{i=1}^{n}(i+1)^2 = \frac{1}{6}n\left(2n^2+9n+13\right) \text{ for } n \in N,$

where $\sum_{i=1}^{n}(i+1)^2 = 2^2+3^2+4^2+\ldots+(n+1)^2$.