more of an algebra problem...
In the identity,
(x+2).....(x+n)} = \sum_{k=0}^n \frac{A_{k}}{x+k})
prove that ^{k}\binom{n}{k})
Thanks in advance(Happy)
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)
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
is many ways to satisfy the first formula.
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

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.
Re: more of an algebra problem...
Quote:
Originally Posted by
earthboy
In the identity,
(x+2).....(x+n)} = \sum_{k=0}^n \frac{A_{k}}{x+k})
prove that
Thanks in advance(Happy)
My way is to,
from
,
put
to get
Is this ok?
can you guys post other solutions..
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
(x+2)......(x+k-1)(x+k+1)...(x+n))
,
put

to get
^{k}k!(n-k)!=n! )
This makes sense.