Problem - 3152
Show that $\sin{x}+2\sin{2x}+\cdots + n\sin{nx}=\frac{(n+1)\sin{nx} - n\sin{(n+1)x}}{2(1-\cos{x})}$
Let $S=\sin{x}+2\sin{2x}+\cdots + n\sin{nx}$. Then
\begin{align*}
&2\cos{x}\cdot S \\
&= \sin{2x} + 2(\sin{3x}+\sin{x}) + \cdots + n(\sin{(n+1)x} + \sin{(n-1)x})\\
&= 2\cdot S + n\sin{(n+1)x}-(n+1)\sin{nx}
\end{align*}
$$\therefore\quad S=\frac{(n+1)\sin{nx}-n\sin{(n+1)x}}{2(1-\cos{x})}$$