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## Main Question or Discussion Point

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

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- Thread starter Suk-Sci
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- #1

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

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- #2

HallsofIvy

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The "base case", of course, is q= 0:

[tex]\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)[/tex]

Now, assume that

[tex]\sum_{m=0}^{k} (n-m) \frac{(p-m)!}{m!} = \frac{(p+k+1)!}{k!} (\frac{n}{p+1} - \frac{k}{p+2} )[/tex]

Then

[tex]\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)!}[/tex]

so you need to show that

[tex] \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)[/tex]

- #3

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

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