# 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!!

• \cdots and \implies ($\cdots$ and $\implies$) might be useful for further questions. – JnxF Jan 24 '16 at 0:09

\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}