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The sum $$g_k=\sum_{j,i:i+j=k} (-1)^j\left(\begin{array}{c} m\\j\end{array}\right) \left(\begin{array}{c} n\\i\end{array}\right),$$ for $0\leq j\leq m$ and $0\leq i \leq n$, is involved in the development of the polynomial $(1+x)^n(1-x)^m$. It seems to me that the sum $$\sum_k\frac{g_k}{k+1}=\frac{2^{n+m}}{(n+m+1)\left(\begin{array}{c} n+m\\n\end{array}\right)}.$$ 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
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2 Answers

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 $(m+1,n+1)$.

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Thank you, as for the post above, it answers perfectly to the question. –  Antonio Sciarretta Mar 31 '12 at 16:48
This cannot be, since the two answers give different results (as gracefully acknowledged by @Steven). –  Did Apr 20 '12 at 16:21
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On the one hand, $$ \int_0^1 \sum_k g_k x^k \,dx = \left.\sum_k \frac{g_k x^{k+1}}{k+1}\right|_0^1 = \sum_k \frac{g_k}{k+1} $$ On the other, by substituting $u=\tfrac12 (1+x)$ and using the symmetry of the integrand [edit: see below] we get $$\int_0^1 \sum_k g_k x^k \,dx = \int_0^1 (1+x)^n (1-x)^m \,dx = 2^{n+m+1} \int_{1/2}^1 u^n (1-u)^m \,du \\ = 2^{n+m} \int_0^1 u^n (1-u)^m \,du = 2^{n+m} B(n+1,m+1) $$ which is what you wanted. (Here $B(\cdot,\cdot)$ is the Beta function.) (The sums over $k$ are formally infinite but actually finite because $(g_k)_{k\in\mathbb Z}$ has finite support; so no issues of convergence arise.)

Edit: Didier quite rightly points out that the integrand isn't symmetric. See Didier's answer for a correct version of this idea.

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I see that Davide Giraudo gave this answer in a comment, more concisely. Also, I forgot to mention: the key word for searching for material on this kind of technique is "generating function". –  Steven Taschuk Mar 31 '12 at 16:25
My comments was more concise since I didn't do the computations. –  Davide Giraudo Mar 31 '12 at 16:31
The integrand is not symmetric when $n\ne m$. –  Did Mar 31 '12 at 16:33
Thank you, perfect and clear. I still wonder anyway if $g_k$ could be explicitly calculated, i.e., without summation –  Antonio Sciarretta Mar 31 '12 at 16:44
Perfect and clear but wrong, though, as @Didier pointed out. (As for a closed form for $g_k$, I don't know of one.) –  Steven Taschuk Mar 31 '12 at 16:46
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