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

$= ((1 + i x)^{2})^{2 n - 1}$

$= (1 - x^{2} + 2 i x)^{2 n - 1}$

$= [(1 - x^{2}) + i (2 x)]^{2 n - 1}$
Total number of terms will be $2 n - 1 + 1 = 2 n$ .
Hence the number of real negative terms will therefore be
$= \frac{2 n}{2}$
$= n$ .

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The number of real negative terms in the binomial expansion of (1+ix)4n−2, nϵN, x>0, is

The correct option is A n(1+ix)4n−2=((1+ix)2)2n−1=(1−x2+2ix)2n−1=[(1−x2)+i(2x)]2n−1Total number of terms will be 2n−1+1=2n.Hence the number of real negative terms will therefore be =2n2=n.

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

The correct option is A $n$
$(1 + i x)^{4 n - 2}$

$= ((1 + i x)^{2})^{2 n - 1}$

$= (1 - x^{2} + 2 i x)^{2 n - 1}$

$= [(1 - x^{2}) + i (2 x)]^{2 n - 1}$
Total number of terms will be $2 n - 1 + 1 = 2 n$ .
Hence the number of real negative terms will therefore be
$= \frac{2 n}{2}$
$= n$ .