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I found that the following equation holds for integers $l$, $k$, and any $x \neq 0,1$,

$\tag{1} \sum\limits_{l = 0}^k {\left( { - 1} \right)^l } \left( {\begin{array}{*{20}c} k \\ l \\ \end{array}} \right)\frac{{x^{l} }}{{\left( {\frac{{1 + l}}{x} + k - l} \right)\left( {\frac{l}{x} + 1 + k - l} \right)}} = \frac{x\left( {1 - x} \right)^{k}}{{k + 1}} $

both in numerically by Matlab and analytically by Mathematica.

So I think there is a reference proving the equation. I have searched equaitons in Wolfram, Wiki, and some tables of series, But I couldn't find any related one.

Actually there were some equations looks like this $\tag{2} \sum\limits_{l = 0}^k {\left( { - 1} \right)^l } \left( {\begin{array}{*{20}c} k \\ l \\ \end{array}} \right)f(k,l,x) = g(k,x), $

but no help.

Also I tried in this way: break the equation into two terms like this

$\tag{3} \sum\limits_{l = 0}^k {\left( { - 1} \right)^l } \left( {\begin{array}{*{20}c} k \\ l \\ \end{array}} \right)\frac{{x^{l} }}{{\left( {\frac{{1 + l}}{x} + k - l} \right)}} + \sum\limits_{l = 0}^k {\left( { - 1} \right)^l } \left( {\begin{array}{*{20}c} k \\ l \\ \end{array}} \right)\frac{{x^{l} }}{{\left( {\frac{l}{x} + 1 + k - l} \right)}}, $

and ran it in Mathematica. But they result in Gauss hypergeometric functions, $_2 F_1 (-k,*,*,x)$, with some coefficients, respectively.

Also $\times2$, I tried to prove it by myself showing the equation holds for $k=0$ and any $x \neq0,1$, then it holds as well when $k+1$ by using the eq (1). but I couldn't...beacuse the $k+1$ case becomes a totally different equation compared to eq (1)...... lol

How can I prove it or find a proof?

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    Thanks for your advice. It was really impressive to me. So I tried and got a resamble equation as follows: $ (lhs)=\sum\limits_{l = 0}^k {\left( { - 1} \right)^l \left( {\begin{array}{*{20}c} k \\ l \\ \end{array}} \right)} \frac{{x^l }}{{\left( {\frac{{k + 1}}{x}} \right)}} $ which leads to one possiblility that $ (k+1)/x = {\left( {\frac{{1 + l}}{x}+k-l}\right)\left({\frac{l}{x}+1+k-l} \right)}$ for any integer $l \ge 0$. But this equality does not hold even $l=0$. I understood, of course from your advice, those summation equations are of the same average but different variances...2012-08-03

1 Answers 1

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We have, $\frac{1}{\left(\frac lx + k - l+1 \right) \left(\frac{l+1}{x} + k-l \right)} = \frac{x}{1-x} \left(\frac{1}{\frac lx + k - l+1} - \frac{1}{\frac{l+1}{x} + k-l} \right)$

Therefore, $\sum_{l=0}^{k} (-1)^l \binom{k}{l} \frac{x^l}{\left(\frac lx + k - l+1 \right) \left(\frac{l+1}{x} + k-l \right)} = \frac{x}{1-x} \left(\sum_{l=0}^{k} (-1)^l \binom{k}{l} \frac{x^l}{\frac lx + k - l+1} - \sum_{l=0}^{k} (-1)^l \binom{k}{l} \frac{x^l}{\frac{l+1}{x} + k-l} \right)$

Call the term inside the bracket as $S$. So, we only need to show that $S = \displaystyle \frac{(1-x)^{k+1}}{k+1}$.

Now, $\displaystyle \sum_{l=0}^{k} (-1)^l \binom{k}{l} \frac{x^l}{\frac lx + k - l+1} = \frac 1{k+1} + \sum_{l=0}^{k-1} (-1)^{l+1} \binom{k}{l+1} \frac{x^{l+1}}{\frac {l+1}x + k - l} $, and

$\displaystyle \sum_{l=0}^{k} (-1)^l \binom{k}{l} \frac{x^l}{\frac {l+1}x + k - l} = \frac {(-1)^k x^{k+1}}{k+1} + \sum_{l=0}^{k-1} (-1)^{l} \binom{k}{l} \frac{x^{l}}{\frac {l+1}x + k - l} $

Thus, on subtracting, we get that

$S = \frac{1}{k+1} + \frac {(-x)^{k+1}}{k+1} + \sum_{l=0}^{k-1} \left((-1)^{l+1} \binom{k}{l+1} \frac{x^{l+1}}{\frac {l+1}x + k - l} + (-1)^{l+1} \binom{k}{l} \frac{x^{l}}{\frac {l+1}x + k - l}\right)$

Now, since $\binom{k}{l+1} = \frac{k-l}{l+1} \binom{k}{l}$, therefore,

$(-1)^{l+1} \binom{k}{l+1} \frac{x^{l+1}}{\frac {l+1}x + k - l} + (-1)^{l+1} \binom{k}{l} \frac{x^{l}}{\frac {l+1}x + k - l} = (-1)^{l+1} \binom{k}{l} \frac{x^{l+1}}{l+1 +(k-l)x} \left(\frac{k-l}{l+1} x + 1\right) = (-1)^{l+1} \binom{k}{l} \frac{x^{l+1}}{l+1} = (-1)^{l+1} \binom{k+1}{l+1} \frac{x^{l+1}}{k+1}$

Thus, $\displaystyle S = \sum_{l=0}^{k+1} \binom{k+1}{l} \frac{x^l}{k+1} = \frac{(1-x)^{k+1}}{k+1}$, which was what we needed.

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    I deeply appriciate your answer. I've never thought that it can be solved in this way. I followed up your equations and it was great pleasure. Thanks2012-08-08