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!!

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

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

  • $\begingroup$ And then I plug in the original value for the answer, or can I just leave it be? $\endgroup$ – EDEDE Jan 23 '16 at 23:55
  • $\begingroup$ @EDEDE see edition. $\endgroup$ – Jane Jan 23 '16 at 23:57
  • $\begingroup$ Cheers, thanks! $\endgroup$ – EDEDE Jan 23 '16 at 23:58

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