Proof via induction $1\cdot3 + 2\cdot4 + 3\cdot5 + \cdots + n(n+2) = \frac{n(n+1)(2n+7)}{6}$ (b) Prove that for every integer $n \ge 1$, $$1\cdot3 + 2\cdot4 + 3\cdot5 + \cdots + n(n+2) = \frac{n(n+1)(2n+7)}{6}$$
This is the second part of a two part question.  Part (a) was the following: 
Write the sum: $1\cdot3 + 2\cdot4 + 3\cdot5 + \cdots + n(n+2)$ using summation notation. 
It was simple enough :  $\sum  k(k+2)$.
For this question, the base case $(n=1)$ holds, as $1\cdot(1+2) = 3 = (1\cdot2\cdot9)/6$.
Induction step: Assume the above holds for all $n = k$, prove that it holds for all $n = k+1$
I'm a bit lost from here, help?

Begin with $$\frac{n(n+1)(2n+7)}{6} = \sum^n_{i=1}n(n+2)$$ and then add $(n+1)(n+3)$ to both sides and try to rewrite the left side in form  $$\frac{(n+1)(n+2)(2(n+1)+7)}{6}.$$
