<s>[INST] Summing $r(r+3)$ using induction We want to prove the following summation by induction:
$$\sum_{r=1}^{n}r(r+3)=\frac{1}{3}n(n+1)(n+5)$$
The problem is posted for a friend, but others can look at the solution if they want/need.
 [/INST] Let us denote the sum by $S_{n}$. We first establish the base case, that $S_{1}=1(1+3)=4$ is equal to $\frac{1}{3}\cdot 1\cdot 2\cdot 6=4$, which holds.  
Then, assume that for some $k$, that $S_{k}=\frac{1}{3}k(k+1)(k+5)$. Obviously $S_{k+1}=S_{k}+(k+1)(k+4).$ Tidying up, we have
$$S_{k+1}=\frac{1}{3}(k+1)\left[k^{2}+5k+3k+12\right]=\frac{1}{3}(k+1)(k^{2}+8k+12)=\frac{1}{3}(k+1)(k+2)(k+6)$$
Which is as we expected. If $S_{k}$ is given by the formula above, then $S_{k+1}$ is. Therefore, since the claim holds for $n=1$, it holds for all natural $n$. $\square$
</s>