# Thread: more of an algebra problem...

1. ## more of an algebra problem...

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

2. ## 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)

3. ## 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 $A_k$ is many ways to satisfy the first formula.

4. ## Re: more of an algebra problem...

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 $A_k$ 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.

5. ## Re: more of an algebra problem...

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

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

Is this ok?
can you guys post other solutions..

6. ## Re: more of an algebra problem...

Ah, so the first equality is true for all x.

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