Question

The value of $i-i^2+i^3-i^4+....-i^{100}$ is equal to

• A. $i$
• B. $-i$
• C. $1-i$
• D. $1+i$
• E. 0

Answer 1

Answer: (E)

$i-i^2+i^3-i^4+...-i^{100}$
This form a GP with common ratio
$(r=-1)$ . $S_n=\frac{i\{1-{(-i)}^{100}\}}{1-(-i)}$
$=\frac{i(1-i^{100})}{1+i}$
$=\frac{i(1-1)}{1+i}$ $(\because i^{100}=1)$
$\therefore$ $S_n=0$