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I'm looking for a proof of this identity:

$$ 1 = \sum_{k=0}^n (-1)^k { 2n \choose n,k,n-k } \frac{n}{n+k} $$

I'll take anything, but a combinatorial proof would be nice - all of the terms in the sum appear to be integers.

Update: Given J.M.'s reformulation, if we start with $$ x^{n-1} (1-x)^n = \sum_{k=0}^n { n \choose k } (-1)^k x^{n+k-1} $$ and integrate both sides from 0 to 1 wrt $x$ we get: $$ \int_0^1 x^{n-1} (1-x)^n dx = \sum_{k=0}^n { n \choose k } \frac{(-1)^k}{n+k} $$ and so it is sufficient to prove that the integral is $1/( n { 2n \choose n } )$.

My instinct tells me to try a trigonometric substitution ($x = \cos^2 u$?) to evaluate the integral - haven't worked out all the details, though. (Update: see leslie townes comment below.)

In any case, I would really like to find a combinatorial proof.

Update 2: Found this paper: Walking into an absolute sum and the sum I'm interested in is $P_n(1)$ where $P_n(x)$ is the polynomial defined by: $$ P_0(x) = 1 \\ P_{n+1}(x) = x^2 [ P_n(x) - P_n(x-1) ] + x P_n(x-1) $$ From this definition it is clear that $P_n(0) = 0$ for $n > 0$ and so $P_n(1) = 1$.

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    Welcome to math.se! What have you tried? What do you know? Answering these types of questions can help us help you.2012-05-09
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    Would $$1=n\binom{2n}{n}\sum_{k=0}^n \binom{n}{k}\frac{(-1)^k}{n+k}$$ be easier for you to prove?2012-05-09
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    That's what I originally started with! I'll take another look...2012-05-09
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    For evaluating the integral, considering the properties of the family of integrals in http://en.wikipedia.org/wiki/Beta_function may help.2012-05-09
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    Yes - that will work!2012-05-09

3 Answers 3