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I have to evaluate this expression: $\sum_k\binom{n}{k}\binom{2k}{k}(-2)^{-k}$,

(In the original question we had $\sum_k\binom{n}{k}\binom{2k}{k}(-2)^{k}$)

this is what I have done:

$$\begin{aligned} \sum_n\sum_k\binom{n}{k}\binom{2k}{k}(-2)^{-k}x^n & = \sum_k\sum_n\binom{n}{k}\binom{2k}{k}(-2)^{-k}x^n \\ & = \sum_k\binom{2k}{k}(-2)^{-k}\sum_n\binom{n}{k}x^n \\ & = \sum_k\binom{2k}{k}(-2)^{-k}\frac{x^k}{(1-x)^{k+1}} \\ & = \frac{1}{1-x}\sum_k\binom{2k}{k}(\frac{x}{2x-2})^k \end{aligned}$$

now we know that $\sum_k\binom{2k}{k}x^k=\frac{1}{\sqrt{1-4x}}$ and so we get

$$ \sum_n\sum_k\binom{n}{k}\binom{2k}{k}(-2)^{-k}x^n = \frac{1}{\sqrt{1-x^2}} $$

So the sum that I'm looking for is equal to $\sum_{k=0}^n\binom{-1/2}{k}(-1)^k\binom{-1/2}{n-k}$

there is a way to express this expression without using sums or menus signs?

share|improve this question
    
There is a typo $(-2)^k \frac{x^k}{(1-x)^{k+1}} = \frac{(-2 x)^k}{(1-x)^{k+1}}$, not $\frac{1}{1-x} \left( \frac{-2x}{2x-2} \right)^k$. –  Sasha Nov 2 '11 at 5:51
    
@Sasha: I fixed it –  Alex M Nov 2 '11 at 6:03
    
Some people may not know what "Snake Oil" means. See section 4.3 of Wilf, generatingfunctionology: math.upenn.edu/~wilf/DownldGF.html –  Michael Lugo Nov 2 '11 at 14:47

1 Answer 1

Added: This post answers the original question, which is to compute $\sum_{k=0}^\infty \binom{n}{k} \binom{2k}{k} (-2)^k$.


Assuming the typo is corrected you correctly arrived at the result: $$ \sum_{n=0}^\infty c_n x^n = \frac{1}{\sqrt{(1-x)(1+7 x)}} $$ Where $c_n = \sum_{k=0}^\infty \binom{n}{k} \binom{2k}{k} (-2)^k$. Using $$ \frac{1}{\sqrt{1-x}} = \frac{1}{\sqrt{\pi}} \sum_{n=0}^\infty \frac{\Gamma(n+1/2)}{n!} x^n \qquad \frac{1}{\sqrt{1+7x}} = \frac{1}{\sqrt{\pi}} \sum_{n=0}^\infty \frac{\Gamma(n+1/2)}{n!} (-7 x)^n $$ Thus $$ \frac{1}{\sqrt{(1-x)(1+7x)}} = \sum_{n=0}^\infty x^n \sum_{m=0}^n \frac{1}{\pi} (-7)^m \frac{\Gamma(m+1/2) \Gamma(n-m+1/2)}{m! (n-m)!} $$

The sum over $m$ is the closed form which can, alternatively, be represented as a hypergeometric polynomial, giving $$ \begin{eqnarray} c_n &=& \sum_{m=0}^n \frac{1}{\pi} (-7)^m \frac{\Gamma(m+1/2) \Gamma(n-m+1/2)}{m! (n-m)!} \\ &=& \frac{\Gamma\left(n+\frac{1}{2}\right)}{\Gamma\left(\frac{1}{2}\right) n!} {}_2 F_1\left( \frac{1}{2}, -n; \frac{1}{2}-n ; -7 \right) \\ &=& {}_2 F_1 \left( \frac{1}{2}, -n ; 1 ; 8 \right) \end{eqnarray} $$


Added: To address OP's skepticism that the above is the correct solution to the question posed, here is some numerical verification using Mathematica:

In[73]:= Table[{Sum[
   Binomial[n, k] Binomial[2 k, k] (-2)^k, {k, 0, n}], 
  Hypergeometric2F1[1/2, -n, 1, 8], (
  Gamma[1/2 (1 + 2 n)] Hypergeometric2F1[1/2, -n, 1/2 - n, -7])/(
  Sqrt[\[Pi]] Gamma[1 + n])}, {n, 1, 6}]

Out[73]= {{-3, -3, -3}, {17, 17, 17}, {-99, -99, -99}, {609, 609, 
  609}, {-3843, -3843, -3843}, {24689, 24689, 24689}}
share|improve this answer
    
It should be simpler, we didn't study the gamma function and the meaning of the symbol $_2F_1$ –  Alex M Nov 2 '11 at 7:15
2  
@AlexM Have you checked that you copied the problem correctly? –  Phira Nov 2 '11 at 7:52
    
Actually the professor told me that there is a typo in the exercise, I will edit the question –  Alex M Nov 4 '11 at 2:40
    
we should have $(-2)^{-k}$ instead of $(-2)^k$ –  Alex M Nov 4 '11 at 2:48
    
@AlexM You should not have removed the previous edition of your question, but rather updated the question, so that readers of this post could relate what I wrote to your question. Please make sure that the original question is available. –  Sasha Nov 4 '11 at 3:50

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