Question

The value of $\sum _{n=1}^{13}(i^n+i^{n+1}),i=\sqrt{-1}$, is

• A. $i$
• B. $i-1$
• C. $1$
• D. $0$

Answer 1

Answer: (B)

$i-1$

$\sum _{n=1}^{13}(i^n+i^{n+1})=\sum _{n=1}^{13}{i}^n(1+i)=(1+i)\sum _{n=1}^{13}{i}^n$

$=(1+i)[i+i^2+i^3+.............+i^{13}]$

$=(1+i)\cdot \frac{i(1-i^{13})}{1-i}$, sum of G.P.

$=(1+i)\cdot \frac{i(1-i)}{1-i}$, since $i^{13}=i^{12}\cdot i=1\cdot i=i$

$=i(1+i)=i-i^2=i-1$, since $i^2=-1$