Generating a non-Wolfram problem

Consider the expansion:

$\displaystyle \frac{1}{\sqrt{1-4x}}=\sum_{k=0}^{\infty} {2k \choose k}x^k$

Replace $x$ with $-x$ and add the two expressions to get:

$\displaystyle \sum_{k=0}^{\infty} {4k \choose 2k} x^{2k}=\frac{1}{2}\left(\frac{1}{\sqrt{1-4x}}+\frac{1}{\sqrt{1+4x}}\right)$

Integrate with respect to $x$ with the limits $0$ to $x$ and divide both the sides by $x$ to obtain:

$\displaystyle \sum_{k=0}^{\infty} \frac{{4k \choose 2k}}{2k+1}x^{2k}=\frac{\sqrt{1+4x}-\sqrt{1-4x}}{4x}$

Substitute $x=1/4$ to get the answer.

Note that WolframAlpha cannot even solve the problem!

One must realize that this hints at a generating function: $\sqrt {\dfrac {1 - \sqrt {1-x}}{x}} = \dfrac {1}{\sqrt{2}} \cdot \displaystyle\sum_{k=0}^{\infty} \dfrac {(4k)!}{16^k \cdot (2k)! (2k+1)!} \cdot x^k.$ Substituting $x = 1$ , we get $\text {LHS} = 1$ , so the answer is $\boxed{\sqrt{2}}$ .

I would have solved this the right way, but because of the title, I decided to do this:

sum of ((4k)!/(16^k * (2k)! * (2k+1)!)) from k=0 to infinity

Put that into W|A and it will yield the answer.