What is the value of the sum ∑_ n=1 ^ 20 (i^ n-1 +i^n+i^ n+1 ) where i= -1 ?

Correct answer: B. 0 aligned & Suppose A=∑_ n=1 ^ 20 (i^ n-1 +i^n+i^ n+1 ) & A=∑_ n=1 ^ 20 i^n ·((1)/(i)+1+i)=∑_ n=1 ^ 20 i^n(-i+1+i) & =∑_ n=1 ^ 20 i^n=i^1+i^2+i^3+i^4 ..+i^ 20 & =5 ·(i+i^2+i^3+i^4) & =5 ·(0)=0\ i+i^2+i^3+i^4=0\ aligned

NDA 2024

Exam: NDA
Year: 2024
Subject: Mathematics
Chapter: Complex Number

NDA 2024 · Mathematics · Complex Number

What is the value of the sum ∑_ n=1 ^ 20 (i^ n-1 +i^n+i^ n+1 )...

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}?

Options

A $-2 i$
B $0$
C $1$
D $2 i$

Correct answer

B. $0$

Step-by-step 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}

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