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$\begingroup$

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

I thought of solving it by induction but am unable to fathom which variable should I apply induction principle upon. I thought of applying on q, but I am stuck. Please help.

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    If we take $q=1$ then the left hand is $n\times p! + (n-1)\times (p+1)!$ and the right hand is $n\times p! \times (p+2) - (p+1)!$, not equal in general. Or have I missed something? – 2017-02-26
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    Let me check, wait a min. please – 2017-02-26
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    My mistake, never mind. Those two expressions actually are equal, though they certainly don't look it. – 2017-02-26
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    Yes, now you are correct, I almost got confused, those are same when simplified! – 2017-02-26
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    Please would uou help me with an answer? – 2017-02-26
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    Check out what I posted. Proof is by induction on $q$. The algebra is simple, but messy and error prone so I recommend checking it carefully. It should work out. – 2017-02-26
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    Let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/54371/discussion-between-amitayas-banerjee-and-lulu). – 2017-02-26
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    Sorry, I have to run out. Does the argument look solid? Granted it doesn't convey much insight. – 2017-02-26

1 Answers 1

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We'll proceed by induction on $q$. The statement can be verified for $q=1$ (though as the comments show, it takes a little effort). Suppose it is known up to $q-1$, let's address it for $q$.

The left hand is, by induction, $$\frac {(p+q)!}{(q-1)!}\times \left(\frac n{p+1}-\frac {q-1}{p+2}\right)+(n-q)\frac {(p+q)!}{q!}$$

Grouping the terms involving $n$ we get $$n\times \frac {(p+q+1)!}{q!(p+1)}$$

Grouping the terms without $n$ we get $$-\frac {(p+q+1)!}{(q-1)!}\times \frac 1{p+2}$$

Adding these together quickly reduces to the right hand.

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    Got it, but is it impossible to assume it true for q and prove for q+1? – 2017-02-26
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    That't the same. just replace $q$ with $q+1$ everywhere in my argument. Good way to check that I didn't mess the algebra up anywhere. – 2017-02-26