The number of real negative terms in the bionomial expansion of $(1 + i x)^{4 n - 2}, n \in N, x > 0$ is

n

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n + 1

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n - 1

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2 n

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Solution

The correct option is A $n$
${(1 + i x)}^{4 x - 2} = \sum_{r = 0}^{^{4 x - 2}}^{^{4 x - 2}} C_{r} {(i x)}^{r}$
For real negative terms, we have to get $i^{2};$ Since $(i^{2} = - 1) .$ For that, $r$ must be equal to $4 k - 2,$ where $k \in I$
Hence, the total real negative terms in the expansion $= n$
$p . s .$ $i = \sqrt{- 1}$
$\Rightarrow$ $i^{2} = - 1$
$\Rightarrow$ $i^{3} = - \sqrt{- 1} = - i$
$\Rightarrow$ $i^{4} = {(i^{2})}^{2} = (- 1)^{2} = 1$
Hence, the answer is $n .$

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