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Let $n\in\mathbb{N}$. For $0\le l\le n$ consider \begin{equation} b_l:=4^{-l} \sum_{j=0}^l \frac{\binom{2 l}{2 j} \binom{n}{j}^2}{\binom{2 n}{2 j}}\text{.} \end{equation} Do you know a technique how to prove that \begin{equation} b_l\ge b_n\text{,$\quad 0\le l\le n-1$?} \end{equation} Going through a long list of binomial identities I did not find epiphany.

Addition: Plot of $b_l$ for $n=20$. Plot of <span class=$b_l$ for $n=20$">

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    May be $b_l$ can be interpreted as probability of some event.2012-08-26
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    precarious: Why/How do you know the result holds?2012-08-26
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    @did Numerical experiments gave sufficient evidence.2012-08-26
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    @precarious I have a solution, I hope it is correct.2012-08-26
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    @precarious before giving any answer do you have any simulation result for $l=10^6,...,10^7$?2012-08-26
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    precarious: I am not sure you answered my question.2012-08-26
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    @SeyhmusGüngören: I do not have it yet. Maxima is calculating. I guess that Mathematica would be much faster.2012-08-26
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    @did: Could you please be more precise?2012-08-26
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    precarious: Well, you certainly did not wake up one morning saying to yourself: "Hey, today let me prove that [complicated sum of fractions of products of binomial coefficients] is greater than [other complicated sum of fractions of products of binomial coefficients]"... There is a reason why you are interested in these sums and there is a reason why you believe they behave the way you ask us to prove they behave--but, sadly, you mentioned none.2012-08-26
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    @did: I think that your assessment is unfair. Of course, there is a reason why I am interested in the problem. Your original question, however, is different and completely answered above.2012-08-26
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    *I think that your assessment is unfair*. OK. You might mention which part(s) is (are) unfair and why. *Of course, there is a reason why I am interested in the problem*. OK. You might wish to mention this reason. If you do not want to, just say so. *Your original question, however, is different and completely answered above*. Certainly not. (And let us leave it at that, shall we.)2012-08-26

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