Find the sum: $\sum_{i=1}^{n}\dfrac{1}{4^i\cdot\cos^2\dfrac{a}{2^i}}$ Find the sum of the following :
$S=\dfrac{1}{4\cos^2\dfrac{a}{2}}+\dfrac{1}{4^2\cos^2\dfrac{a}{2^2}}+...+\dfrac{1}{4^n\cos^2\dfrac{a}{2^n}}$
 A: Hint:
$$\frac{1}{4^n\cos^2\frac{a}{2^n}}+\frac{1}{4^n\sin^2\frac{a}{2^n}}=
\frac{1}{4^{n-1}\sin^2\frac{a}{2^{n-1}}}.$$

Adding $\displaystyle \frac{1}{4^n\sin^2\frac{a}{2^n}}$ to the sum, the result thus telescopes to $\displaystyle\frac{1}{\sin^2a}$, and hence the initial sum is
$$\frac{1}{\sin^2a}-\frac{1}{4^n\sin^2\frac{a}{2^n}}.$$
A: A very different approach to calculate the sum. Note that
$$
\frac{1}{4^k\cos^2\frac{x}{2^k}}=\left(-\ln\cos\frac{x}{2^k}\right)''.
$$
Let's calculate (using in the $3^\text{d}$ equality the general form of Morrie's law for $\alpha=\frac{x}{2^n}$)
$$
F(x)=\sum_{k=1}^n-\ln\cos\frac{x}{2^k}=-\ln\prod_{k=1}^n\cos\frac{x}{2^k}=-\ln\frac{\sin x}{2^n\sin\frac{x}{2^n}}=\ln\sin\frac{x}{2^n}-\ln\sin x+n\ln 2.
$$
Finally, our sum is
$$
F''(a)=\frac{1}{\sin^2a}-\frac{1}{4^n\sin^2\frac{a}{2^n}}.
$$
A: $$S=S+\frac{1}{4^n \sin^2\frac{a}{2^n}}-4^{-n}\csc^2 (a/2^n)\\=\sum_{k=1}^{n-1}\frac{1}{4^k\cos^2(a/2^k)}+\frac{1}{4^{n-1}\sin^2{a/2^{n-1}}}-4^{-n}\csc^2 (a/2^n)\\
=\cdots=\frac{1}{\sin^2(a)}-4^{-n}\csc^2 (a/2^n)=\csc^2(a)-4^{-n}\csc^2 (a/2^n)$$
