The sum of

We're going to use the identity $\sin(2x) = 2 \sin(x) \cos(x)$ and telescoping.

Defining $a_n$ to be $\begin{aligned} a_{n} &= \left( 2 \sin^4 \frac{\pi}{2^n} - 6 \cos^2 \frac{\pi}{2^n} \sin^2 \frac{\pi}{2^n} \right) \\ &= \left( 2 \sin^2 \frac{\pi}{2^n} \left( \sin^2 \frac{\pi}{2^n} - 3 \cos^2 \frac{\pi}{2^n} \right) \right) \\ &= \left( 2 \sin^2 \frac{\pi}{2^n} \left( 1 - 4 \cos^2 \frac{\pi}{2^n} \right) \right) \\ &= \left( 2 \sin^2 \frac{\pi}{2^n} - 8 \cos^2 \frac{\pi}{2^n} \sin^2 \frac{\pi}{2^n} \right) \\ &= \left( 8 \cos^2 \frac{\pi}{2^{n+1}} \sin^2 \frac{\pi}{2^{n+1}} - 8 \cos^2 \frac{\pi}{2^n} \sin^2 \frac{\pi}{2^n} \right) \end{aligned}$

Therefore, the summation up to $k^{th}$ term will be $\begin{aligned} S_k &= \sum_{n=1}^{k} \left( 8 \cos^2 \frac{\pi}{2^{n+1}} \sin^2 \frac{\pi}{2^{n+1}} - 8 \cos^2 \frac{\pi}{2^n} \sin^2 \frac{\pi}{2^n} \right) \\ &= \sum_{n=2}^{k+1} \left( 8 \cos^2 \frac{\pi}{2^n} \sin^2 \frac{\pi}{2^n} \right) - \sum_{n=1}^{k} \left( 8 \cos^2 \frac{\pi}{2^n} \sin^2 \frac{\pi}{2^n} \right) \\ &= 8 \cos^2 \frac{\pi}{2^{k+1}} \sin^2 \frac{\pi}{2^{k+1}} - 8 \cos^2 \frac{\pi}{2} \sin^2 \frac{\pi}{2} \end{aligned}$

As $\cos \frac{\pi}{2} = 0$ and $\lim_{k\to\infty} \sin^2 \frac{\pi}{2^{k+1}} = 0$ , we can deduct $\lim_{k\to\infty} S_k = \mathcal{S} = 0$