If $n$ is odd, $^{n} {C_{1} +}^{n} {C_{3} +}^{n} {C_{5} + . . . . . . . . +}^{n} C_{2 [n / 2] - 1 =}$ where $[.]$ denotes greatest integer function

2^{n - 1}

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2^{n - 1} - 1

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2^{n}

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2^{n} - 1

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Solution

The correct option is B $2^{n - 1} - 1$
$we know that$
$if n \in odd$
$\Rightarrow n = 2 r + 1 -(i) (where r is an integer)$
$\Rightarrow \frac{n}{2} = \frac{2 r + 1}{2} = r + \frac{1}{2}$

$\Rightarrow [\frac{n}{2}] = [r + \frac{1}{2}] = [r] + [\frac{1}{2}]$

$\Rightarrow [\frac{n}{2}] = r + 0$

$\Rightarrow [\frac{n}{2}] = \frac{n - 1}{2}$

from eqn. (i)

So, $^{n} c_{1} +^{n} c_{3} +^{n} c_{5} +^{n} c_{7} + \dots . . +^{n} C_{2 [\frac{n}{2}] - 1}$

$=^{n} c_{1} +^{n} c_{3} +^{n} c_{5} +^{n} c_{7} + \dots +^{n} c_{n - 2}$

$n \in$ odd

$\Rightarrow (1 + x)^{n} =^{n} c_{0} +^{n} c_{1} x +^{n} c_{2} x^{2} +^{n} c_{3} x^{3} + \dots$

$+^{n} c_{n - 1} x^{n - 1} +^{n} c_{n} x^{n}$

on putting $x = 1$ we get,

$2^{n} = n_{c_{0}} +^{n} c_{1} +^{n} c_{2} +^{n} c_{3} + \dots +^{n} c_{n - 1} +^{n} c_{n}$

we know that $n_{C_{r}} = n_{C_{n - r}}$

$\Rightarrow 2^{n} =^{n} c_{n} +^{n} c_{1} +^{n} c_{n - 2} +^{n} c_{3} +^{n} c_{n - 4} +^{n} c_{5} + \dots .$

$+^{n} c_{n - 2} +^{n} c_{1} +^{n} c_{n}$

$\Rightarrow 2^{n} = 2 [n_{c_{1}} +^{n} c_{3} +^{n} c_{5} +^{n} c_{7} + \dots +^{n} c_{n - 2} +^{n} c_{n}]$

$\Rightarrow \frac{2^{n}}{2} =^{n} c_{1} +^{n} c_{3} +^{n} c_{5} +^{n} c_{7} + \dots +^{n} c_{n - 2} + 1$

$\Rightarrow 2^{n - 1} - 1 =^{n} c_{1} +^{n} c_{3} +^{n} c_{5} + \dots \cdot^{n} c_{n - 2}$

$\Rightarrow^{n} c_{1} +^{n} c_{3} +^{n} c_{5} + \dots \dots^{n} c_{2 [\frac{n}{2} - 1]} = 2^{n - 1} - 1$

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