Prove that $n^3+3n^2+3n+1=2(1+2+...+n)(n+1)+(n+1)^2$ I'm trying to prove that $n^3+3n^2+3n+1=2(1+2+...+n)(n+1)+(n+1)^2$, which is part of a larger proof. We can write:
$n^3+2n^2+n=2(1+2+...+n)(n+1)$
$n^3+2n^2+n=(1+2+...+n)(2n+2)$
$n^3+2n^2+n=(2n+4n+6n+...+2n^2)+(2+4+6+...+n)$
I have no idea how to proceed, though.

You can simplify by $n+1$ to get
$$(n+1)^2=2(1+2+\cdots n)+(n+1).$$
Then by induction,
$$(n+1)^2-n^2=2n+1=2(1+2+\cdots n)-2(1+2+\cdots n-1)+(n+1)-n.$$
