Would anyone be able to assist me with this proof by Induction?

$$\sum\_{p=1}^n p(p+2) = \frac{n(n+1)(2n+7)}{6}$$

Here's my attempt at a solution (skipping the base case step):

1) Assume true for n = k:
$$\sum\_{p=1}^kp(p+2) = \frac{k(k+1)(2k+7)}{6}$$

2) Add k + 1st term to both sides:
$$k(k+2) + (k+1)(k+1+2) = \frac{k(k+1)(2k+7)}{6} + \frac{k+1(k+1+1)(2(k+1)+7)}{6}$$

3) Inductive step:
$$\frac{k(k+1)(2k+7)}{6} + (k+1)(k+1+2) = \frac{k(k+1)(2k+7)}{6} + \frac{k+1(k+1+1)(2(k+1)+7)}{6}$$

Is the setup correct thus far? I have never been able to prove equality but could be making algebraic mistakes.

Thanks!