sequences and series - Finding $Sigmafrac{(n+1)(n+2)}{2!}$ .

How to find the summation of $\Sigma\frac{(n+1)(n+2)}{2!}$ ?

MY WORK:

I know that the expression in the summation is the general term of the binomial expansion $(1-x)^{-3}$ . I have a solution where I consider :
$$(1-x)^{-3}=a_0+a_1x+a_2x^2 +...+a_nx^n +...$$
Then I multiply it with the following :
$$(1-x)^{-1}=1+x+x^2+x^3+ ... + x^n+ ...$$
Equating co efficients of $x^n$ , I get :
$$a_0+a_1+a_2+...+a_n=\Sigma\frac{(n+1)(n+2)}{2!}=\frac{(n+1)(n+2)(n+3)}{3!}$$ ...

CAN I DO IT IN ANY OTHER WAY OTHER THAN BINOMIAL EXPANSION?

Answer

You can do it like this: $$\begin{align}\sum_{k=0}^n\frac{(k+1)(k+2)}2&=\frac12\sum_{k=0}^nk^2+\frac32\sum_{k=0}^nk+\sum_{k=0}^n1\\&=\frac{n(n+1)(2n+1)}{12}+\frac{3n(n+1)}4+n+1.\end{align}$$ Now, all it takes is to check that $$\frac{n(n+1)(2n+1)}{12}+\frac{3n(n+1)}4+n+1=\frac{(n+1)(n+2)(n+3)}6.$$