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 $$2[(\tan^2x+1)(3\tan^2x+1)+(\tan^22x+1)(3\tan^22x+1)…]$$ 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 $$log(cosx\cos(2x)\cos(2^2x)...\cos(2^nx))=log(\frac{\sin(2^{n+1}x)}{2^{n+1}sinx})$$ 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.

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.

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 $$2[(\tan^2x+1)(3\tan^2x+1)+(\tan^22x+1)(3\tan^22x+1)…]$$ 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 $$log(cosx\cos(2x)\cos(2^2x)...\cos(2^nx))=log(\frac{\sin(2^{n+1}x)}{2^{n+1}sinx})$$ 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.

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 $$2[(\tan^2x+1)(3\tan^2x+1)+(\tan^22x+1)(3\tan^22x+1)…]$$ 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 $$log(cosx\cos(2x)\cos(2^2x)...\cos(2^nx))=log(\frac{\sin(2^{n+1}x)}{2^{n+1}sinx})$$ 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.

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 $$2[(\tan^2x+1)(3\tan^2x+1)+(\tan^22x+1)(3\tan^22x+1)…]$$ 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 $$log(cosx\cos(2x)\cos(2^2x)...\cos(2^nx))=log(\frac{\sin(2^{n+1}x)}{2^{n+1}sinx})$$ so series of $\tan^2$ gets solved but how to approach series of $\tan^4$?

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 $$2[(\tan^2x+1)(3\tan^2x+1)+(\tan^22x+1)(3\tan^22x+1)…]$$ 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 $$log(cosx\cos(2x)\cos(2^2x)...\cos(2^nx))=log(\frac{\sin(2^{n+1}x)}{2^{n+1}sinx})$$ 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.