The value of $(- I)^{1 + 3 + 5 + . . . . . + (2 n + 1)}$ is

I

for all values of n

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- I

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0

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None

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Solution

The correct option is D None
$I^{1 + 3 + 5 + . . . (2 n + 1)}$
$= I^{(n + 1)^{2}}$
$= I^{n^{2} + 2 n + 1}$
Let $n = 2 k$
$I^{4 k^{2} + 4 k + 1}$
$I^{4} (k^{2} + k) + 1$
$= I^{4} (k^{2} + k) \times I$
$= 1 \times I$
$= I$
If $n$ is odd, $n + 1$ will be even
Let $n + 1 = 2 m$
$∴ I^{(n + 1)^{2}} = I^{(2 m)^{2}} = I^{4 m^{2}}$
$= 1$

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