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7.Binomial Theorem
medium
$\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 $n$
B
$\frac{{{2^{n - 1}}}}{{n!}}$; for all values of $n$ i.e., all even odd values
C
$0$
D
None of these
Solution
(b) Multiplying each term by $n!$ the question reduces to
$\frac{{n!}}{{1!(n – 1)!}} + \frac{1}{{3!}}.\frac{{n!}}{{(n – 3)\,!}} + \frac{1}{{5!}}.\frac{{n!}}{{(n – 5)!}} + ….$
$ = {\,^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}}$.
Standard 11
Mathematics
