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Would anyone be able to assist me with this proof by Induction?

$$\sum_{p=1}^n p(p+2) = \frac{n(n+1)(2n+7)}{6}$$

Here's my attempt at a solution (skipping the base case step):

1) Assume true for n = k: $$\sum_{p=1}^kp(p+2) = \frac{k(k+1)(2k+7)}{6}$$

2) Add k + 1st term to both sides: $$k(k+2) + (k+1)(k+1+2) = \frac{k(k+1)(2k+7)}{6} + \frac{k+1(k+1+1)(2(k+1)+7)}{6}$$

3) Inductive step: $$\frac{k(k+1)(2k+7)}{6} + (k+1)(k+1+2) = \frac{k(k+1)(2k+7)}{6} + \frac{k+1(k+1+1)(2(k+1)+7)}{6}$$

Is the setup correct thus far? I have never been able to prove equality but could be making algebraic mistakes.

Thanks!

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    $\begingroup$ Why not split it into $$\sum_{k=1}^nk+k(k+1)=\frac{n(n+1)}2+\frac{n(n+1)(n+2)}3$$ which is found from the hockey-stick identity. $\endgroup$ – Simply Beautiful Art Dec 8 '16 at 0:58
  • $\begingroup$ The last step is wrong... $\endgroup$ – Simply Beautiful Art Dec 8 '16 at 1:00
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At the end, you should just have

$$(k+1)(k+1+2)=\frac{(k+1)[6(k+3)]}6$$

Add in $k(k+1)(2k+7)/6$ to get

$$\frac{(k+1)[k(2k+7)]}6+\frac{(k+1)[6(k+3)]}6=\frac{(k+1)[2k^2+7k+6k+18]}6\\=\frac{(k+1)(k+2)(2k+9)}6\\=\frac{(k+1)(k+1+1)(2(k+1)+7)}6$$

and we are done!

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  • $\begingroup$ Thanks! How did you come up with $$\frac{(k+1)[6(k+3)]}{6}$$ on the RHS? $\endgroup$ – Sean Dec 8 '16 at 1:18
  • $\begingroup$ @Sean Notice that I multiplied it by $\frac66$, or$$x=\frac{6x}6$$ $\endgroup$ – Simply Beautiful Art Dec 8 '16 at 1:20
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The assumption is wrong. You have to assume that

$$\sum_{k=1}^n k(k+2) = \frac{n(n+1)(2n+7)}{6}$$

and add the $n+1$'st term to the both sides:

$$\sum_{k=1}^{n+1} k(k+2) = \frac{n(n+1)(2n+7)}{6} + (n+1)(n+3)$$

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  • $\begingroup$ I see no real difference between your suggestions and the OP's steps. I imagine it is just a typo. $\endgroup$ – Simply Beautiful Art Dec 8 '16 at 1:05

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