Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $g_{k}$ ?,

share|improve this question
    
Yes, thanks, I have corrected my question –  Antonio Sciarretta Mar 31 '12 at 16:02
1  
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 at 5:35

1 Answer 1

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

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.