Finding $\Sigma\frac{(n+1)(n+2)}{2!}$ .

Question

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 1

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$$\frac{n(n+1)(2n+1)}{12}+\frac{3n(n+1)}4+n+1=\frac{(n+1)(n+2)(n+3)}6.$$

Answer 2

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

Answer 3

Under the convention that for nonnegative integers $r,k$: $$\binom{k}{r}=0\text{ if }k<r$$ it can be proved by induction that: $$\sum_{k\leq n}\binom{k}{r}=\binom{n+1}{r+1}$$

Induction step:$$\sum_{k\leq n}\binom{k}{r}=\sum_{k\leq n}\binom{k-1}{r-1}+\sum_{k\leq n}\binom{k-1}{r}=\binom{n}{r}+\binom{n}{r+1}=\binom{n+1}{r+1}$$

For $r=2$ we find: $$\sum_{k=0}^{n}\frac{(k+1)(k+2)}2=\sum_{k=0}^{n}\binom{k+2}{2}=\sum_{k\leq n+2}\binom{k}{2}=\binom{n+3}{3}=\frac{(n+1)(n+2)(n+3)}6$$