Write $\sum_{k=1}^nk\sin(kx)^2$ in closed form

Question

$\underline{Given:}$

Write in closed form $$\sum_{k=1}^nk\sin(kx)^2$$ using the fact that $$\sum_{k=1}^nku^k=\frac u{(1-u)^2}[(n)u^{n+1}(n+1)u^n+1]$$

$\underline{My\ Work:}$

I substituted $\sin(kx)^2$ for $\frac{1-\cos(2kx)}2$ which let me rewrite the summation like this: $$\frac 12\bigg[\sum_{k=1}^nk-\sum_{k=1}^nk\cos(2kx)\bigg]$$

I could rearrange the summation again using the fact that $\cos(2kx)=\frac {e^{i2xk}+e^{-i2xk}}2$

$$\frac 14\bigg[n(n+1)-\sum_{k=1}^nk({e^{i2x}})^k-\sum_{k=1}^nk({e^{i2x}})^k\bigg]$$ Then I re-wrote the summation again using the substitution that was given to me. Using $u=e^{i2x}$ $$\frac 14\bigg[n(n+1)-\frac{e^{i2x}}{(1-e^{i2x})^2}\bigg[n(e^{i2x})^{n+1}(n+1)(e^{i2x})^n+1\bigg]-\frac{e^{-i2x}}{(1-e^{-i2x})^2}\bigg[n(e^{-i2x})^{n+1}(n+1)(e^{-i2x})^n+1\bigg]$$

Which if I multiply the $\frac{e^{-i2x}}{(1-e^{-i2x})^2}$ by $\frac {e^{4ix}}{e^{4ix}}$ it changes into $\frac{e^{i2x}}{(1-e^{i2x})^2}$ which allows me to factor it out.

Now I have: $$\frac 14\bigg[n(n+1)-\frac{e^{i2x}}{(1-e^{i2x})^2}\bigg[\big[n(e^{i2x})^{n+1}(n+1)(e^{i2x})^n+1\big]-\big[n(e^{-i2x})^{n+1}(n+1)(e^{-i2x})^n+1\big]\bigg]$$

Now this is where I'm starting to get stuck. I can't seem to organize this nicely to get back into sin's and cos's. Any help would be appreciated! Thanks!

Answer 1

We have $\dfrac{e^{2ix}}{(1-e^{2ix})^2}=\dfrac{-1}{4sin^2(x)}$ and $n(e^{2ix})^{n+1}(n+1)(e^{2ix})^n-n(e^{-2ix})^{n+1}(n+1)(e^{-2ix})^n=2in(n+1)sin(4n+2)$.