Q10 / 120 · NDA 2024 (Phase 2) · Mathematics

What is the value of the sum $\sum_{n=1}^{20}\left(i^{n-1}+i^n+i^{n+1}\right)$ where $i=\sqrt{-1}$ ?

  • A
    $-2 i$
  • Correct answer
    B
    $0$
  • C
    $1$
  • D
    $2 i$

Solution

$\begin{aligned} & \text { Suppose } A=\sum_{n=1}^{20}\left(i^{n-1}+i^n+i^{n+1}\right) \\ & \Rightarrow A=\sum_{n=1}^{20} i^n \cdot\left(\frac{1}{i}+1+i\right)=\sum_{n=1}^{20} i^n(-i+1+i) \\ & =\sum_{n=1}^{20} i^n=i^1+i^2+i^3+i^4 \ldots \ldots . .+i^{20} \\ & =5 \cdot\left(i+i^2+i^3+i^4\right) \\ & =5 \cdot(0)=0\left\{\because i+i^2+i^3+i^4=0\right\}\end{aligned}$

Hence, the correct answer is option: 'B'.

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