(D) Given expression: $i^n + i^{n+1} + i^{n+2} + i^{n+3}$ Factor out $i^n$: $i^n(1 + i + i^2 + i^3)$ We know that $i^2 = -1$ and $i^3 = -i$. Substitut

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(D) Given expression: $i^n + i^{n+1} + i^{n+2} + i^{n+3}$ Factor out $i^n$: $i^n(1 + i + i^2 + i^3)$ We know that $i^2 = -1$ and $i^3 = -i$. Substituting these values: $i^n(1 + i - 1 - i)$ Simplifying the expression inside the parentheses: $i^n(0) = 0$ Thus, the sum of any four consecutive powers of $i$ is $0$.

Class 11 Mathematics 1 Complex Numbers and Quadratic Equations: Complex Numbers Integral power of iota, Algebraic operations and Equality of complex numbers

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If $i = \sqrt{-1}$ and $n$ is a positive integer, then $i^n + i^{n+1} + i^{n+2} + i^{n+3}$ is equal to

WBJEE 2009 All WBJEE

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$1$
B
$i$
C
$i^n$
D
$0$

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1 Complex Numbers and Quadratic Equations: Complex Numbers

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Integral power of iota, Algebraic operations and Equality of complex numbers

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If $i = \sqrt{-1}$ and $n$ is a positive integer, then $i^n + i^{n+1} + i^{n+2} + i^{n+3}$ is equal to

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