In the identity,

$\displaystyle \frac{n!}{x(x+1)(x+2).....(x+n)} = \sum_{k=0}^n \frac{A_{k}}{x+k}$

prove that $\displaystyle A_{k}=(-1)^{k}\binom{n}{k}$

Thanks in advance(Happy)

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- Jul 5th 2012, 03:24 AMearthboymore of an algebra problem...
In the identity,

$\displaystyle \frac{n!}{x(x+1)(x+2).....(x+n)} = \sum_{k=0}^n \frac{A_{k}}{x+k}$

prove that $\displaystyle A_{k}=(-1)^{k}\binom{n}{k}$

Thanks in advance(Happy) - Jul 5th 2012, 10:45 AMpanda89Re: more of an algebra problem...
using mi is the way to go.

for n=0, lhs = rhs = 1/x

assume it holds for n

prove it still holds for n+1, using this hint 1/[(x+k)(x+n+1)] = [ 1/(x+k) - 1/(x+n+1) ] / (n+1) - Jul 6th 2012, 12:58 AMemakarovRe: more of an algebra problem...
What does it mean to prove something "in the identity"? Do you mean to prove that the first identity implies the second one? That would be surprising because I would expect one can choose $\displaystyle A_k$ is many ways to satisfy the first formula.

- Jul 6th 2012, 02:52 AMearthboyRe: more of an algebra problem...
- Jul 6th 2012, 03:01 AMearthboyRe: more of an algebra problem...
- Jul 6th 2012, 03:17 AMemakarovRe: more of an algebra problem...