661. Consider, instead, the following sum:
S=e^{ix}+2e^{i2x}+3e^{i3x}+\ldots + n e^{inx}
The required sum in the problem is, then, the real part of S. Multiply S with eix and subtract from S to obtain
(1-e^{ix})S=e^{ix}+e^{i2x}+e^{i3x}+\ldots + e^{inx} -ne^{i(n+1)x}
\Rightarrow\ (1-e^{ix})S=e^{ix}\dfrac{1-e^{inx}}{1-e^{ix}} -ne^{i(n+1)x}
As such we obtain,
S=e^{ix}\dfrac{1-e^{inx}}{(1-e^{ix})^2} -\dfrac{ne^{i(n+1)x}}{1-e^{ix}}
Taking the real part and simplifying, we get the required sum
\sum_{k=1}^nk\cos kx = \dfrac{n}{2}\ \dfrac{\sin(n+\frac{1}{2})x}{\sin\frac{x}{2}}-\dfrac{1}{2}\ \dfrac{\sin^2\frac{nx}{2}}{\sin^2\frac{x}{2}}
