(C) We know that $\sqrt{-1} = i$.Therefore, the expression becomes $\sum_{n=1}^{4} (i)^{2n} = i^2 + i^4 + i^6 + i^8$.Since $i^2 = -1$, $i^4 = 1$, $i^6

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(C) We know that $\sqrt{-1} = i$.Therefore, the expression becomes $\sum_{n=1}^{4} (i)^{2n} = i^2 + i^4 + i^6 + i^8$.Since $i^2 = -1$, $i^4 = 1$, $i^6 = -1$, and $i^8 = 1$.Substituting these values: $-1 + 1 - 1 + 1 = 0$.

Class 11 Mathematics 4-1.Complex numbers Integral power of iota, Algebraic operations and Equality of complex numbers

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$\sum_{n=1}^{4} (\sqrt{-1})^{2n} = $

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$2$
B
$-i$
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$0$
D
$i$

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

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$\sum_{n=1}^{4} (\sqrt{-1})^{2n} = $

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