MCQ
$\frac{1}{{1!(n - 1)\,!}} + \frac{1}{{3!(n - 3)!}} + \frac{1}{{5!(n - 5)!}} + .... = $
- A$\frac{{{2^n}}}{{n!}}$; for all even values of $n$
- ✓$\frac{{{2^{n - 1}}}}{{n!}}$; for all values of $n$ i.e., all even odd values
- C$0$
- DNone of these
$\frac{{n!}}{{1!(n - 1)!}} + \frac{1}{{3!}}.\frac{{n!}}{{(n - 3)\,!}} + \frac{1}{{5!}}.\frac{{n!}}{{(n - 5)!}} + ....$
$ = {\,^n}{C_1} + {\,^n}{C_3} + {\,^n}{C_5} + .... = {2^{n - 1}}$.
Thus $\frac{1}{{1!(n - 1)!}} + \frac{1}{{3!(n - 3)!}} + \frac{1}{{5!(n - 5)!}} + ....$$ = \frac{1}{{n!}}{2^{n - 1}}$.
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