proof of a finite sum involving a binomial coefficient and a variable.

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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i woult try to take term $(1-x)^k$ on the right and expand it using the binomial formula. Further more you could write the coefficient $x$ in the denominator (like on the lhs) $\frac{(1-x)^k}{\frac{k+1}{x}}$. Then you can copare the coefficients and see if it works... – vanguard2k Aug 3 '12 at 8:25
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... – Heejin Joung Aug 3 '12 at 11:24

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. Thanks – Heejin Joung Aug 8 '12 at 6:20