<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] $$3r^2+3r=3r(r+1)$$ and $$n(n+1)(n+2)=\dfrac{1}{4}n(n+1)(n+2)[(n+3)-(n-1)]\\=\dfrac{1}{4}n(n+1)(n+2)(n+3)-\dfrac{1}{4}(n-1)n(n+1)(n+2)$$
</s>