Question

The value of ${\sum }_{k=0}^n(i^k+i^{k+1}),$ where $i^2=-1,$ is equal to :

• A. $i-i^n$
• B. $-i+i^{n+1}$
• C. $i-i^{n+1}$
• D. $i-i^{n+2}$
• E. $-i-i^n$

Answer 1

Answer: (D)

$i-i^{n+2}$

The value of ${\sum }_{k=0}^n(i^k+i^{k+1})={\sum }_{k=0}^n{i}^k\left(i+1\right)$ is,
$=(i+1)\left[\frac{1(1-i^{n+1})}{(1-i)}\right]\times \frac{1+i}{1+i}$
$=\frac{(1+i{)}^2}{1+1}=\frac{(1-1+2i)(1-i^{n+1})}{2}$
$=\frac{2i}{2}(1-i^{n+1})$

$=i-i^{n+2}$