Question

Sum upto $100$ term of the series $i+2i^2+3i^3+\cdots$ is

• A. $50(1-i)$
• B. $25i$
• C. $25(1+i)$
• D. $100(1-i)$

Answer 1

The correct option is A $50(1-i)$

$S=i+2i^2+3i^3+\cdots 100i^{100}\cdots \left(1\right)$
$Si=i^2+2i^3+\cdots +99i^{100}+100i^{101}\cdots \left(2\right)$
Subtracting Equation $\left(2\right)$ from equation $\left(1\right),$ we get
$S(1-i)=i+i^2+i^3+\cdots i^{100}-100i^{101}$
$\because \text{Sum of n terms of G.P}{S}_n=\frac{a(1-r^n)}{1-r}$
Here, $a=i,r=i,n=100$
$\Rightarrow S(1-i)=\frac{i(1-i^{100})}{1-i}-100i^{100}\cdot i$
$[\because i^{100}=(i^4{)}^{25}=1]$
Now, $S(1-i)=0-100i$
$\Rightarrow S=100\left(\frac{-i}{1-i}\right)$
$\Rightarrow S=100\left(\frac{-i(1+i)}{2}\right)$
$\therefore S=50(1-i)$