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Calculus Level 3

$\sum_{n=1}^\infty \binom{4n}{2n}^{-1}=X+\frac{\pi}{\sqrt{Y}}-\frac{2}{Z\sqrt{Z}}\ln\left(\frac{1+\sqrt{Z}}{2}\right)$

The equation above holds true for rational numbers $X$ , $Y$ , and $Z$ . Find $\sqrt{XYZ}$ .

Note : $\binom{\cdot}{\cdot}$ is the binomial coefficient . The first few terms of the series are as follows:

$\begin{aligned} \sum_{n=1}^\infty \binom{4n}{2n}^{-1} &= \frac{1}{\binom{4}{2}} + \frac{1}{\binom{8}{4}} + \frac{1}{\binom{12}{6}} + \cdots \\ &= \frac{1}{6}+\frac{1}{70}+\frac{1}{924}+\cdots. \end{aligned}$

The answer is 9.

1 solution

Ilya Z.
Feb 4, 2018

The solution could be found here : see Theorem 3.8 on page 11. The last thing to be considered here is to ensure that the lower summation limits are not identical: 1 - here and 0 - via the link formulae.

Hardest Brilliant problem I've seen. You can't even be bothered to give the solution. No, I didn't get it, but I did verify the approximate solution.

The cited page gives X=16/15

pi/sqrt(Y)=pi*sqrt(3)/27 from which Y=243

2/(Z sqrt(Z))=2 sqrt(5)/25 from which Z=5

XYZ=1296

sqrt(1296)=36

Jeremy Galvagni - 3 years, 4 months ago

$X=\dfrac{1}{15}$ . Because In this question the lower summation limit starts from $n=1$ .

Digvijay Singh - 3 years, 4 months ago

There are two approaches that can lead to the exact solution. But for me, both of them requires programs like Wolfram Alpha or Mathetica to finish the last step.

Method 1

${{2n \choose n}^{-1}} = \int_0^1(2n+1)y^n(1-y)^ndy$

With this identity, the answer will be:

$\int_0^1\frac{1 + 3z}{(1 - z)^2}dy - 1, z = (y(1-y))^2$

Method 2,

$f(x) = \sum_{n=0}^\infty {2n \choose n }^{-1}x^n$

Then what we want is $\frac{f(1) + f(-1)}{2} - 1$ .

$f(x) = _2F_1(1, 1; \frac{1}{2}; \frac{x}{4})$

Where $_2F_1$ is the hypergeometric function.

Stimim Chen - 3 years, 3 months ago

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The answer is 9.

1 solution

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