How should i solve : $$\sum_{r=1}^n (2r-1)\cos(2r-1)\theta $$
I can solve $\sum_{r=1}^n cos(2r-1)\theta $ by considering $\Re \sum_{r=1}^n z^{2r-1} $ and using summation of geometric series, but I can't seem to find a common geometric ratio when $ 2r-1 $ is involved in the summation.
Visually : $\sum_{r=1}^n z^{2r-1} = z +z^3+...+z^{2r-1}$ where the common ratio $ r= z^2 $ can easily be seen, but in the case of $\sum_{r=1}^n (2r-1)z^{2r-1} = z + 3z^3 + 5z^5 +...+ (2r-1)z^{2r-1}$, how should i solve this ? A hint would be appreciated.