Hola
The expression \[ \displaystyle \sum^{2017}_{n=0}\bigg[\sin(2^{n+1}x)\prod^{n}_{k=0}\cos^2(2^kx)\bigg] \]
We can start in this way:
\( \displaystyle\prod^{n}_{k=0}cos(2^kx)=cos(x)cos(2x)cos(4x)\ldots cos(2^n x)=\dfrac{1}{sin(x)}sin(x)cos(x)cos(2x)cos(4x)\ldots cos(2^n x)=\\
\\\quad =\dfrac{1}{2sin(x)}sin(2x)cos(2x)cos(4x)\ldots cos(2^nx)=\dfrac{1}{2^2sin(x)}sin(4x)cos(4x)\ldots cos(2^nx)=\\
\\\quad =\ldots=\dfrac{1}{2^{n+1}sin(x)}sin(2^{n+1}x) \)
So:
\( \displaystyle \sum^{2017}_{n=0}\bigg[\sin(2^{n+1}x)\prod^{n}_{k=0}\cos^2(2^kx)\bigg]=\dfrac{1}{sin^2(x)}\displaystyle \sum^{2017}_{n=0}\bigg[\dfrac{1}{4^{n+1}}\sin^3(2^{n+1}x)\bigg] \)
Now I am not sure how we can continue. An idea could be call:
\( f(x)=\displaystyle \sum^{2017}_{n=0}\bigg[\dfrac{1}{4^{n+1}}\sin^3(4^{n+1}x)\bigg] \)
Then:
\( f''(x)=\dots=3\displaystyle \sum^{2017}_{n=0}\bigg[sin(2^{n+2}x)cos(2^{n+1}x)-sin^2(2^{n+1}x))\bigg] \)
But I don't know if this is usefull.
Best regards.
