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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...

Mark S.
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Apoorv
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1 Answers1

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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 $LHS$.

GrzegorzK
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