If $$S_1=2\sum_{k=0}^n 16^k \tan^4 {2^k x}
$$
$$S_2=4\sum_{k=0}^n 16^k \tan^2 {2^k x}
$$
$$S_3= \sum_{k=0}^n 16^k
$$
Find $(3S_1 + 2S_2 + 2S_3)$ as a function of $x$ and $n.$
In the expression asked it rearranges to $$2[(\tan^2x+1)(3\tan^2x+1)+(\tan^22x+1)(3\tan^22x+1)\cdots]$$ Now this is not simplifying further, I tried in converting $\tan$ to $\sec$ but no approach still found. Maybe there is some special series devoted to summation of $\tan^2$ and $\tan^4$ about which I’m not aware.
I also thought differentiating twice the expression $$\log(\cos (x)\cos(2x)\cos(2^2x)\cdots\cos(2^nx))=\log\left(\frac{\sin(2^{n+1}x)}{2^{n+1}\sin(x)}\right)$$ so series of $\tan^2$ can be created by $\tan^2\theta+1=\sec^2\theta$ but the constant terms are coming wrong and series of $\tan^4$ is still far away.
Best Answer
If $x=m\pi\ (m\in\mathbb Z)$, then $$3S_1+2S_2+2S_3=\dfrac{2(16^{n+1}-1)}{15}$$
If $x\not=m\pi\ (m\in\mathbb Z)$, then $$3S_1+2S_2+2S_3=\dfrac{2^{4n+5}(\tan^2(2^{n+1}x)+3)(\tan^2(2^{n+1}x)+1)}{\tan^4(2^{n+1}x)}-\dfrac{2(\tan^2x+3)(\tan^2x+1)}{\tan^4x}$$
Proof :
For $x=m\pi\ (m\in\mathbb Z)$, since $\sin x=0$, we get $$3S_1+2S_2+2S_3=0+0+2S_3=\dfrac{2(16^{n+1}-1)}{15}$$
In the following, $x\not=m\pi\ (m\in\mathbb Z)$.
Let us start with $\displaystyle\int(3S_1+2S_2+2S_3)dx$.
Using $$\begin{align}\int \tan^4(2^kx)dx&=\frac{\tan^3(2^kx)}{3\cdot 2^k}-\frac{\tan(2^kx)}{2^k}+x+C \\\\\int\tan^2(2^kx)dx&=\frac{\tan(2^kx)}{2^k}-x+C\end{align}$$ we have $$\begin{align}&\int(3S_1+2S_2+2S_3)dx \\\\&=\sum_{k=0}^{n}\bigg( 6\cdot 16^k\int \tan^4(2^kx)dx+8\cdot 16^k\int\tan^2(2^kx)dx+2\cdot 16^k\int dx\bigg) \\\\&=\sum_{k=0}^{n}\bigg[6\cdot 16^k\bigg(\frac{\tan^3(2^kx)}{3\cdot 2^k}-\frac{\tan(2^kx)}{2^k}+x\bigg) \\&\qquad\quad +8\cdot 16^k\bigg(\frac{\tan(2^kx)}{2^k}-x\bigg)+2\cdot 16^kx\bigg]+C_1 \\\\&=\sum_{k=0}^{n}\bigg(2\cdot 8^k\tan^3(2^kx)-6\cdot 8^k\tan(2^kx)+6\cdot 16^kx \\&\qquad\quad +8\cdot 8^k\tan(2^kx)-8\cdot 16^kx+2\cdot 16^kx\bigg)+C_1 \\\\&=\sum_{k=0}^{n}\bigg(2\cdot 8^k\tan^3(2^kx)+2\cdot 8^k\tan(2^kx)\bigg)+C_1 \\\\&=\underbrace{\sum_{k=0}^{n}\bigg(\frac{2\cdot 8^k\tan(2^kx)}{\cos^2(2^kx)}\bigg)}_{F_1}+C_1\end{align}$$
Using $$\int\frac{\tan(2^kx)}{\cos^2(2^kx)}dx=\frac{\tan^2(2^kx)}{2^{k+1}}+C$$ we have $$\begin{align}\int F_1\ dx&=\sum_{k=0}^{n}\bigg(2\cdot 8^k\int \frac{\tan(2^kx)}{\cos^2(2^kx)}dx\bigg) \\\\&=\sum_{k=0}^{n} \bigg(2\cdot 8^k\cdot \frac{\tan^2(2^kx)}{2^{k+1}}\bigg)+C_2 \\\\&=\underbrace{\sum_{k=0}^{n}\bigg(2^{2k}\tan^2(2^kx)\bigg)}_{F_2}+C_2\end{align}$$
So, we have $$\begin{align}\int F_2\ dx&=\sum_{k=0}^{n} 2^{2k}\int \tan^2(2^kx)dx \\\\&=\sum_{k=0}^{n} 2^{2k}\bigg(\frac{\tan(2^kx)}{2^k}-x\bigg)+C_3 \\\\&=\underbrace{\sum_{k=0}^{n}\bigg(2^k\tan(2^kx)-2^{2k}x\bigg)}_{F_3}+C_3\end{align}$$
Using $$\int \tan(2^kx)dx=-\frac{\log|\cos(2^kx)|}{2^k}+C$$ we have $$\begin{align}\int F_3\ dx&=\sum_{k=0}^{n}\bigg(2^k\int\tan(2^kx)dx-2^{2k}\int xdx\bigg) \\\\&=\sum_{k=0}^{n}\bigg(-\log|\cos(2^kx)|-2^{2k-1}x^2\bigg)+C_4 \\\\&=-\sum_{k=0}^{n}\log|\cos(2^kx)|-x^2\sum_{k=0}^{n}2^{2k-1}+C_4 \\\\&=-\log\bigg|\frac{\sin(2^{n+1}x)}{2^{n+1}\sin x}\bigg|-\frac{2^{2n+2}-1}{6}x^2+C_4 \\\\&=\underbrace{-\log|\sin(2^{n+1}x)|+\log|\sin x|-\frac{2^{2n+2}-1}{6}x^2}_{F_4}+C_5\end{align}$$
We can get $3S_1+2S_2+2S_3$ by differentiating $F_4$ four times, so $$3S_1+2S_2+2S_3=\dfrac{2^{4n+5}(\tan^2(2^{n+1}x)+3)(\tan^2(2^{n+1}x)+1)}{\tan^4(2^{n+1}x)}-\dfrac{2(\tan^2x+3)(\tan^2x+1)}{\tan^4x}.\ \blacksquare$$
There is another solution using the idea of telescoping sum.
Letting $$G(k):=\frac{2^{4k+1}(\tan^2(2^{k}x)+3)(\tan^2(2^{k}x)+1)}{\tan^4(2^{k}x)}$$ we have $$\begin{align}3S_1+2S_2+2S_3&=\sum_{k=0}^{n}2\cdot 16^k(3\tan^2(2^kx)+1)(\tan^2(2^kx)+1) \\\\&=\sum_{k=0}^{n}\bigg(G(k+1)-G(k)\bigg) \\\\&=G(n+1)-G(0)\end{align}$$