<s>[INST] Prove by induction $\sum_{i=1}^n 3i(i+4) = \frac{(n)(n+1)(2n+13)}{2}$ In this problem I have to prove the following equation using mathematical induction:
$$\sum_{i=1}^n 3i(i+4) = \frac{(n)(n+1)(2n+13)}{2}$$
So far I've proved that $P_1$ is true, and written $P_k$ as $$15+36+63+...+3k(k+4)= \frac{(k)(k+1)(2k+13)}{2}$$
I then attempt to solve $P_{k+1}$. I start by using $P_k$ and substituting it into the equation to get $$\frac{(k)(k+1)(2k+13)}{2}+(3k+1)(k+5)=\frac{(k+1)(k+2)(2k+14)}{2}$$
I then multiply $(3k+1)(k+5)$ by $\frac{2}{2}$ to get $$\frac{(k)(k+1)(2k+13)(3k+1)(2k+10)}{2}=\frac{(k+1)(k+2)(2k+14)}{2}$$
Here its obvious that the LHS isnt going to be equal to the RHS, seeing as the LHS would have $k^5$ amd the RHS would have $k^3$. Any ideas?
 [/INST] There are three silly mistakes here: First off, your expression on the RHS should be $\frac{(k+1)(k+2)(2k+15)}{2}$ since. 
Secondly you somehow made the addition into multiplication here
$$\frac{(k)(k+1)(2k+13)}{2}+(3k+1)(k+5)=\frac{(k+1)(k+2)(2k+14)}{2}$$
And finally as Bernard said you should have a $3(k+1)=3k+3$ instead of $3k+1$ in your inductive step.
Try again starting from here 
$$\frac{(k)(k+1)(2k+13)}{2}+(3k+3)(k+5)=\frac{(k+1)(k+2)(2k+15)}{2}$$
and comment if you need more help.
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