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Q. The value of $\frac{1}{1 !(n-1) !}+\frac{1}{3 !(n-3)}+\frac{1}{5 !(n-5) !}+\ldots \ldots .$. is

Binomial Theorem

Solution:

Let $S=\frac{1}{1 !(n-1) !}+\frac{1}{3 !(n-3)}+\frac{1}{5 !(n-5) !}+\ldots \ldots$
$=\frac{1}{n !}\left(\frac{n !}{1 !(n-1) !}+\frac{n !}{3 !(n-3) !}+\frac{1}{5 !(n-5) !}+\ldots ..\right)$
$=\frac{1}{n !}\left({ }^n C_1+{ }^n C_3+{ }^n C_5+\ldots \ldots\right)$
$=\frac{1}{n !} \cdot \frac{2^n}{2} \Rightarrow \frac{2^{n-1}}{n !}$