Question

The sum of $i-2-3i+4+...$upto $100$ terms, where$i=\sqrt{-1}$ is

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

Answer 1

The correct option is A

$50\left(1-i\right)$

Explanation for correct option

We know that $i^2=-1$ and $i^4=1$

Therefore,$S_n$ the sum of the first $n$ term of the series is

$$\begin{gathered} S_n=i-2-3i+4+....+100i^{100} \\ =1i+2i^2+3i^3+..+100i^{100} \\ \therefore i{S}_n=i^2+2i^3+3i^4+...+100i^{101} \\ \therefore S_n-i{S}_n=\left(i+i^2+i^3+..+i^{100}\right)-100i^{101} \\ \Rightarrow \left(1-i\right)S_n=\frac{i\left(i^{100}-1\right)}{i-1}-100i^{101} \\ \Rightarrow \left(1-i\right)S_n=\frac{i\left(1-1\right)}{i-1}-100i^{101}\left[\because i^{100}=1\right] \\ \Rightarrow \left(1-i\right)S_n=-100i^{101} \\ \Rightarrow S_n=\frac{-100i^{101}}{\left(1-i\right)} \\ \Rightarrow S_n=\frac{-100i^{100}·i·\left(1+i\right)}{\left(1-i\right)\left(1+i\right)} \\ \Rightarrow S_n=\frac{-100i^{100}·\left(i+i^2\right)}{2} \\ \Rightarrow S_n=50·\left(1-i\right)\left[\because i^{100}=1i^2=-1\right] \end{gathered}$$

Hence, option A is correct.