# How to prove $\sum\limits_{k=0}^{A} C_{A+B}^{B+k}C_{B+k-1}^{k}(-1)^{k}=1$?

I am trying to prove this equation directly, but so far without success (unless $B=1$).

$$\sum_{k=0}^{A} C_{A+B}^{B+k}C_{B+k-1}^{k}(-1)^{k}=1$$

I found it can be transformed into beta function, as

$$\int_0^1x^{B-1}(1-x)^{A}dx=\frac{A!(B-1)!}{(A+B-1)!}$$

but is there any other more straight forward way to prove it?

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Is $C_{m}^n$ a binomial coefficient often denoted as $\binom{m}{n}$ and having value $\frac{m!}{n!(m-n)!}$? – Dilip Sarwate Dec 30 '12 at 20:51
@DilipSarwate Yes, you are right. – Ziqian Xie Dec 30 '12 at 20:53
I found it can be transformed into beta function... Does this mean that you are able to prove this identity holds, using a beta identity? Or that you tried to prove it using a beta identity but failed? – Did Dec 30 '12 at 23:05
@did I am able to prove it that way, but I am trying to find another method. – Ziqian Xie Dec 30 '12 at 23:19

Rewriting ${A+B\choose B+k}$ as ${A+B\choose A-k}$ and ${B+k-1\choose k}(-1)^k$ as ${-B\choose k}$, the sum on the LHS becomes $$\sum_{k\geqslant0}{A+B\choose A-k}{-B\choose k},$$ which is the $x^A$ term in the expansion of the product $$(1+x)^{A+B}\cdot(1+x)^{-B}.$$ Since this product is also $(1+x)^A$, the $x^A$ term in its expansion is ${A\choose A}=1$.