$\sum_{k=1}^{n-1}(n-k) \cos \frac{2 k \pi}{n}=____, where $n \geq 3$ is aninteger.\

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Professor PC Thomas Classes · Answer

Let $R=C+i S$\ where $C=\sum_{k=1}^{n-1}(n-k) \cos \frac{2 k \pi}{n}$\ and $S=\sum_{k=1}^{n-1}(n-k) \sin \frac{2 k \pi}{n}$\ $	herefore \quad R=\sum_{k=1}^{n-1}(n-k)\left[\cos \frac{2 k \pi}{n}+i \sin \frac{2 k \pi}{n}ight]$\ $\Rightarrow R=\sum_{k=1}^{n-1}(n-k)\left[\cos \frac{2 \pi}{n}+i \sin \frac{2 \pi}{n}ight]^{k}$\ $\Rightarrow R=\sum_{k=1}^{n-1}(n-k) z^{k}$,\ where $z=\cos \frac{2 \pi}{n}+i \sin \frac{2 \pi}{n}$ $$\begin{equation*}\Rightarrow R=(n-1) z+(n-2) z^{2}+\ldots+z^{n-1} 	ag{i} \end{equation*}$$ $$\begin{equation*}\Rightarrow z R=(n-1) z^{2}+(n-2) z^{3}+\ldots+2 z^{n-1}+z^{n} 	ag{ii} \end{equation*}$$ Subtracting (ii) from (i) we have\ $(1-z) R=(n-1) z-\left[z^{2}+z^{3}+\ldots . .+z^{n}ight]$\ $\Rightarrow(1-z) R=(n-1) z-z^{2}\left[1+z+\ldots+z^{n-2}ight]$\ $\Rightarrow(1-z) R=(n-1) z-z^{2}\left(\frac{1-z^{n-1}}{1-z}ight)$\ $\Rightarrow(1-z) R=(n-1) z-\frac{z^{2}-z^{n} z}{1-z}$\ $\Rightarrow(1-z) R=(n-1) z-\frac{z^{2}-z}{1-z} \quad\left[\because z^{n}=1ight]$\ $\Rightarrow(1-z) R=(n-1) z-\frac{z(z-1)}{(1-z)}$\ $\Rightarrow(1-z) R=(n-1) z+z \Rightarrow R=\frac{n z}{1-z}$\ $\Rightarrow C+i S=\frac{n\left[\cos \frac{2 \pi}{n}+i \sin \frac{2 \pi}{n}ight]}{1-\cos \frac{2 \pi}{n}-i \sin \frac{2 \pi}{n}}$\ $\Rightarrow C+i S=\frac{n\left[\cos \frac{2 \pi}{n}+i \sin \frac{2 \pi}{n}ight]}{-2 i \sin \frac{\pi}{n}\left[\cos \frac{\pi}{n}+i \sin \frac{\pi}{n}ight]}$\ $\Rightarrow C+i S=\frac{-\frac{n}{2}}{\sin \frac{\pi}{n}}\left[\sin \frac{\pi}{n}-i \cos \frac{\pi}{n}ight]$\ Comparing real parts, we have,\ $C=-\frac{n}{2} 	imes \frac{\sin \frac{\pi}{n}}{\sin \frac{\pi}{n}}=-\frac{n}{2}$\