# Solve $\sum_{r=1}^n (2r-1)\cos(2r-1)\theta$

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.

• What about differentiating the first sum and multiplying it by $z$? Feb 24, 2017 at 5:44
• Feb 24, 2017 at 5:45

Note that $$\sum_{r=1}^{n} (2r-1) \cos (2r-1) \theta=\sum_{r=1}^{n} \frac{\mathrm{d}}{\mathrm{d}\theta}\sin (2r-1) \theta= \frac{\mathrm{d}}{\mathrm{d}\theta}\sum_{r=1}^{n} \sin (2r-1) \theta$$ Now calculate $$\sum_{r=1}^{n} \sin (2r-1) \theta=\frac{ \sin^2(n\theta) }{ \sin(\theta) }$$ Through the formula for the sum of sin's.
Now just diffferentiate with regard to $\theta$.
Here's a hint: $${d \over dz} z^n = n z^{n-1}$$. What do you get if you differentiate the geometric series term by term?