Using power series to evaluate another sum I have that $$\frac{1}{\sqrt{1-z}}=\sum_{k=0}^{\infty}\frac{z^k}{2^{2k}}\binom{2k}{k}$$
And I want to use this to evaluate the sum $$\sum_{k=0}^{\infty}\frac{(-1)^k}{2^{6k}}\binom{4k}{2k}$$  I'm just not sure how I would get the binomial coefficients to match, I can see that the numbers are doubled, but is there a neat trick/identity to get it to match the coefficient above.
Thanks
 A: Start by re-writing your sum as follows:
$$\sum_{k\ge 0} \frac{(i/2)^{2k}}{2^{4k}}{4k\choose2k}$$
where $i$ is the imaginary unit.
Now observe that by cancellation of odd powers, we have that
$$\frac{1}{2}\left(\frac{1}{\sqrt{1-z}}+\frac{1}{\sqrt{1+z}}\right) =
\sum_{k\ge 0} \frac{z^{2k}}{2^{4k}}{4k\choose2k}.$$
It follows that your sum is
$$\frac{1}{2}\left(\frac{1}{\sqrt{1-i/2}}+\frac{1}{\sqrt{1+i/2}}\right).$$
Now let $\theta = \arctan(1/2,1)$ and start simplifying, getting
$$\frac{1}{2}
\left(\frac{\sqrt{1+i/2}}{\sqrt{1+1/4}}+\frac{\sqrt{1-i/2}}{\sqrt{1+1/4}}\right)
= \frac{1}{\sqrt{5}}
\left(\sqrt{1+i/2}+\sqrt{1-i/2}\right)\\
=\frac{1}{\sqrt{5}} \sqrt{\frac{\sqrt{5}}{2}} 2\cos(1/2\times\theta).$$
Now we have
$$\cos(1/2\times\theta) = \sqrt{\frac{1+\cos\theta}{2}}.$$
To find $\cos\theta$ consider the triangle with vertices $(0,0)$, $(1,0)$ and $(1,1/2)$. Scale by $\sqrt{5}/2$ to obtain a point on the unit circle, getting $(0,0)$, $(2/\sqrt{5},0)$ and $(2/\sqrt{5},1/\sqrt{5}).$ It is now immediate that $$\cos\theta = 2/\sqrt{5}.$$
This finally yields
$$\cos(1/2\times\theta) = \frac{\sqrt{2}}{2}\sqrt{1+\frac{2}{\sqrt{5}}}$$
which gives for our sum the value
$$\frac{1}{\sqrt{5}} \sqrt{\frac{\sqrt{5}}{2}} \times 2\times
\frac{\sqrt{2}}{2}\sqrt{1+\frac{2}{\sqrt{5}}} =
\sqrt{\frac{2}{5}} \sqrt{1+\frac{\sqrt{5}}{2}}.$$
