Proving summation series

Question

If $m,n,p,q$ are non-negative integers , prove that

$$\sum^{q}_{m=0} (n-m)\frac{(p+m)!}{m!}= \frac{(p+q+1)!}{q!} \left( \frac{n}{p+1}-\frac{q}{p+2} \right)$$

How to do these type of questions

Answer 1

First distribute the fraction on $(n-m)$ as below

$A=\sum^{q}_{m=0} (n-m)\frac{(p+m)!}{m!}=n\sum^{q}_{m=0}\frac{(p+m)!}{m!}-\sum^{q}_{m=0} m\frac{(p+m)!}{m!}$

Then

$A=np!\sum^{q}_{m=0} {{p+m}\choose{p}}-\sum^{q}_{m=0}{{p+m}\choose{p+1}}(p+1)!$

Then you can use a formula, which is

$\sum_{j=k}^n\binom{j}{k}=\binom{n+1}{k+1}$

This formula can be applied to both sums, directly. So, you get

$A=np!{{p+q+1}\choose{p+1}}-(p+1)!{{p+q+1}\choose{p+2}}$

The rest should be easy.