expression for$\sum_{k=0}^{n}(k+1)(k+2)$ what is a closed expression for$\sum_{k=0}^{n}(k+1)(k+2)$? the answer says to look at Generating function $a_k=(k+1)(k+2)$ and derive twice $\dfrac{1}{x}=1+x+x^2+...$ so the solution starts: $\sum_{k=0}^{n}(k+1)(k+2)=[\dfrac{1}{(1-x)}.\dfrac{2}{(1-x)^3}]$but isn't $\sum_{k=0}^{\infty}t^k=\dfrac{1}{(1-t)}$? so how the lhs and rhs in  $\sum_{k=0}^{n}(k+1)(k+2)=[\dfrac{1}{(1-x)}.\dfrac{2}{(1-x)^3}]$ are equal?

$$ \sum_{k=0}^n (k+1)(k+2) = \sum_{k=1}^n k(k+1) = \sum_{k=1}^n k^2+k=$$
$$=\sum_{k=1}^n k^2+\sum_{k=1}^n k = \frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2}=$$
$$=\frac{n(n+1)}{2}\left(\frac{2n+1}{3}+1\right)=\frac{n(n+1)}{2}\frac{2n+4}{3}=\frac{n(n+1)(n+2)}{3}$$
