# more of an algebra problem...

• Jul 5th 2012, 03:24 AM
earthboy
more 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 $\displaystyle A_{k}=(-1)^{k}\binom{n}{k}$

• Jul 5th 2012, 10:45 AM
panda89
Re: 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 AM
emakarov
Re: 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 AM
earthboy
Re: more of an algebra problem...
Quote:

Originally Posted by emakarov
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.

I think the questions says if the first information is given ,we have to prove the second statement.
btw, the first one is is not an identity, and may be the words:"in the identity" are wrong.Its a "if...,then prove" type of problem.
• Jul 6th 2012, 03:01 AM
earthboy
Re: more of an algebra problem...
Quote:

Originally Posted by earthboy
In the identity,
$\displaystyle \frac{n!}{x(x+1)(x+2).....(x+n)} = \sum_{k=0}^n \frac{A_{k}}{x+k}$
prove that $\displaystyle A_{k}=(-1)^{k}\binom{n}{k}$

My way is to,
from $\displaystyle n!=\sum_{k=0}^n A_{k}x(x+1)(x+2)......(x+k-1)(x+k+1)...(x+n)$,
put $\displaystyle x=-k$ to get $\displaystyle A_{k}(-1)^{k}k!(n-k)!=n!$

Is this ok?
can you guys post other solutions..
• Jul 6th 2012, 03:17 AM
emakarov
Re: more of an algebra problem...
Ah, so the first equality is true for all x.

Quote:

Originally Posted by earthboy
My way is to,
from $\displaystyle n!=\sum_{k=0}^n A_{k}x(x+1)(x+2)......(x+k-1)(x+k+1)...(x+n)$,
put $\displaystyle x=-k$ to get $\displaystyle A_{k}(-1)^{k}k!(n-k)!=n!$

This makes sense.