# Sum of products of binomial coefficients: proof

The sum $$g_{k}=\sum_{\vphantom{\LARGE A}j,\,i \atop {\vphantom{\LARGE A}i\ +\ j\ =\ k}} \left(-1\right)^{\,j}{m \choose j}{n \choose i},\qquad \mbox{for}\quad 0 \leq j \leq m\quad\mbox{and}\quad 0 \leq i \leq n,$$ is involved in the development of the polynomial $(1+x)^n(1-x)^m$.

It seems to me that the sum $$\sum_{k}{g_{k} \over k+1} = {2^{n + m} \over (n+m+1){n+m \choose n}}.$$ Could anybody provide a formal proof or a closed-form development of $g_{k}$ ?,

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Yes, thanks, I have corrected my question – Antonio Sciarretta Mar 31 '12 at 16:02
Write $\sum_k\frac{g_k}{k+1}$ as the integral of a product of two polynomials, then use a substitution. You will recognize an equality involving Gamma and Beta functions. – Davide Giraudo Mar 31 '12 at 16:14
The formula is wrong (try $n=1$, $m=2$). – Did Mar 31 '12 at 16:50
@FelixMarin Your edit makes the indices in the first summation nearly unreadable. Is it necessary? I suggest to slow down on the sophistication of the LaTeX encoding (Knuth would be horrified by the... things you use) and to concentrate on the readability of the end result. – Did Jul 4 '14 at 5:35

Writing each $\frac1{k+1}$ as $\int\limits_0^1x^k\mathrm dx$, one gets $S=\int\limits_0^1G(x)\mathrm dx$ with $S=\sum\limits_k\frac{g_k}{k+1}$ and $G(x)=\sum\limits_kg_kx^k$, hence $$G(x)=\sum\limits_{i,j}(-1)^j{m\choose j}{n\choose i}x^{i+j}=\sum\limits_{i}{n\choose i}x^{i}\sum\limits_{j}(-1)^j{m\choose j}x^{j}=(1+x)^n(1-x)^m.$$ Let $x=1-2t$, then $0\leqslant t\leqslant\frac12$, $\mathrm dx=2\mathrm dt$, $1+x=2(1-t)$ and $1-x=2t$, hence $G(x)=2^{n+m}t^m(1-t)^n$ and $$S=2^{n+m+1}\int_0^{1/2}t^m(1-t)^n\mathrm dt=2^{n+m+1}\,\mathrm B_{1/2}(m+1,n+1),$$ where $x\mapsto\mathrm B_x(m+1,n+1)$ is the incomplete Beta function of parameters $(m+1,n+1)$.