The value of the sum $\sum_{n=1}^{13}\left(i^{n}+i^{n+1} ight)$, where $i=\sqrt{-1}$\ equals\

Question

The value of the sum $\sum_{n=1}^{13}\left(i^{n}+i^{n+1}ight)$, where $i=\sqrt{-1}$\ equals\

Professor PC Thomas Classes · Answer

(b) : $\sum_{n=1}^{13}\left(i^{n}+i^{n+1}ight)=\sum_{n=1}^{13} i^{n}(1+i)$\ $=(1+i)\left\{\frac{i\left(1-i^{13}ight)}{1-i}ight\}=(1+i)\left\{\frac{i(1-i)}{1-i}ight\}$\ $=(1+i) i=-1+\mathrm{i}$\