Assistance with proof by Induction

Question

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!

Answer 1

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!

Answer 2

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:

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