# Arithmetic of integer (based on mathematical induction)

If $$m,n,p,q$$ are non-negative integers, prove that $$\sum_{m=0}^{q}(n-m)\frac{(p+m)!}{m!}=\frac{(p+q+1)!}{q!}\left (\frac{n}{p+1}-\frac{q}{p+2}\right )$$ I tried for this but wasn't able to come up with an effective approach. So, please help me...

Use induction on $$q$$:
For $$q = 0$$ the equality holds. Assume that the equality holds for $$q = t-1$$ (with $$t \geq 1$$), we need to show that the equality holds for $$q = t$$.
We have: $$LHS = \sum_{m = 0}^t (n-m) \frac{(p+m)!}{m!}= \sum_{m = 0}^{t-1} (n-m) \frac{(p+m)!}{m!} + (n-t)\frac{(p+t)!}{t!}$$. By induction hypothesis this becomes: $$\frac{(p+t)!}{(t-1)!}(\frac{n}{p+1} - \frac{t-1}{p+2}) + (n-t)\frac{(p+t)!}{t!}$$. After some calculation (making all fractions have the same denominator), we get: $$\frac{(p+t)!}{t!(p+1)(p+2)}(npt +2nt - t^2p -t^2 +pt - t +np^2+3np+2n-tp^2-3tp)$$.
We also have $$RHS = \frac{(p+t+1)!}{t!}(\frac{n}{p+1}-\frac{t}{p+2}) = \frac{(p+t)!}{t!(p+1)(p+2)}(\frac{n}{p+1}-\frac{t}{p+2})(p+1)(p+2)(p+t+1)$$ and it's easy to show that this is equal to $$LHS$$.