<s>[INST] Use the identity $(r+1)^3-r^3\equiv3r^2+3r+1$ to find $\sum_{r=1}^nr(r+1)$ Use the identity $(r+1)^3-r^3\equiv3r^2+3r+1$ to find $$\sum_{r=1}^nr(r+1)$$
I can obtain $$\sum_{r=1}^n3r^2+3r+1=(n+1)^3-1$$
and I think the next step is 
$$3\sum_{r=1}^nr(r^2+1)+\frac13=\left((n+1)^3-1\right)$$
But how do I deal with the constant 3 at the end of the LHS and also tidy up the answer to get $\frac13n(n+1)(n+2)$?
 [/INST] We have
$$r(r+1)=\frac{1}{3}((r+1)^3-r^3-1)$$
so 
$$\sum_{r=1}^n r(r+1)=\frac{1}{3}\sum_{r=1}^n((r+1)^3-r^3)-\frac{1}{3}\sum_{r=1}^n1=\frac{1}{3}((n+1)^3-1)-\frac{n}{3}$$
</s>