Prove Summation: $\sum_{m=0}^{q} (n-m) \frac{(p-m)!}{m!}$

[Suk-Sci]

prove that: \sum_{m=0}^{q} (n-m) \frac{(p-m)!}{m!} = \frac{(p+q+1)!}{q!} (\frac{n}{p+1} - \frac{q}{p+2} ) using induction

 

[HallsofIvy]

The "base case", of course, is q= 0:
\sum_{m=0}^{0} (n-m) \frac{(p-m)!}{m!}= (n-0)\frac{p- 0}{0!}= mp= \frac{(p+ 1)!}{0!}\left(\frac{n}{p+1}\right)

Now, assume that
\sum_{m=0}^{k} (n-m) \frac{(p-m)!}{m!} = \frac{(p+k+1)!}{k!} (\frac{n}{p+1} - \frac{k}{p+2} )

Then
\sum_{m=0}^{k+1} (n-m) \frac{(p-m)!}{m!}= \frac{(p+k+1)!}{k!} (\frac{n}{p+1} - \frac{k}{p+2} )+ (m-k-1)\frac{(p- k- 1)!}{(k+1)!}

so you need to show that
\frac{(p+k+1)!}{k!} (\frac{n}{p+1} - \frac{k}{p+2} )+ (m-k-1)\frac{(p- k- 1)!}{(k+1)!}= \frac{(p+ k+ 2)!}{(k+1)!}\left(\frac{n}{p+1}- \frac{k+1}{p+2}\right)

 

[Suk-Sci]

Thank You...now i have got it ...i made a mistake in m=k+1