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?

Assume result is true for $n=k$ i.e.
$$S_k=\frac {k(k+1)(2k+7)}6$$
Then, adding the next term gives:
$$\begin{align}
S_{k+1}&=S_k+(k+1)(k+2)\\
&=\frac {k(k+1)(2k+7)}6+(k+1)(k+3)\\
&=\frac{k+1}6 \left( k(2k+7)+6(k+3)\right)\\
&=\frac{k+1}6\left( 2k^2+13k+18\right)\\
&=\frac{(k+1)(k+2)(2k+9)}6\\
&=\frac{(\overline{k+1})(\overline{k+1}+1)(2\overline{k+1}+7)}6\\
\end{align}$$
i.e. result is also true for $n=k+1$.
As shown in the question, the result is true for $n=1$. 
Hence, by induction, the result is true for all positive integer $n$.
