Prove: $ 1\times3 +2\times4 + \cdots + n(n+2) = \frac{1}{6} \times n(n+1)(2n+7)$ using Induction I'm told to prove this by Mathematical Induction:
$ 1\times3 +2\times4 + \cdots + n(n+2) = \frac{1}{6} \times n(n+1)(2n+7)$
This is what I have so far:
BC:
Try $n=1$:
$ 1\times3 +2\times4 + \cdots + n(n+2) = \dfrac{1}{6} \times n(n+1)(2n+7)$
$ 3= \dfrac{1}{6} \times 2(9) = 3$
So Base case is true:
IH:
 Let $n = k$.
$ 1\times3 +2\times4 + \cdots + k(k+2) = \frac{1}{6} \times k(k+1)(2k+7)$
IS: show that
$n \implies n+1$
I'm told to only work from one side, so I've attempted the left side (I was told I can't plug this into both sides).
We Have:
$ 1\times3 +2\times4 + \cdots + (k+1)(k+3) = ...$
I'm not sure where to go from here, any help would be greatly appreciated!!

$$
\begin{align}
1 \times 3+2 \times 4+ \cdots +k(k+2)+(k+1)(k+3)
  &=\frac{1}{6} \times k(k+1)(2k+7)+(k+1)(k+3)
\\[5pt]&=(k+1)\left(\frac{2k^2+7k}{6}+k+3\right)
\\[5pt]&=(k+1)\left(\frac{2k^2+7k+6k+18}{6}\right)
\\[5pt]&=(k+1)\left(\frac{2k^2+13k+18}{6}\right)
\\[5pt]&=\frac{(k+1)(k+2)(2(k+1)+7)}{6}
\end{align}
$$
