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$\frac{{{C_0}}}{1} + \frac{{{C_2}}}{3} + \frac{{{C_4}}}{5} + \frac{{{C_6}}}{7} + ....$=
$\frac{{{2^{n + 1}}}}{{n + 1}}$
$\frac{{{2^{n + 1}} - 1}}{{n + 1}}$
$\frac{{{2^n}}}{{n + 1}}$
None of these
Solution
(c) Putting the values of ${C_0},{C_2},{C_4}….,$we get $ = 1 + \frac{{n(n – 1)}}{{3.2!}} + \frac{{n(n – 1)(n – 2)(n – 3)}}{{5.4!}} + ….$
=$\frac{1}{{n + 1}}\left[ {(n + 1) + \frac{{(n + 1)n(n – 1)}}{{3!}} + \frac{{(n + 1)n(n – 1)(n – 2)(n – 3)}}{{5!}} + ….} \right]$
Put $n + 1$=N = $\frac{1}{N}\left[ {N + \frac{{N(N – 1)(N – 2)}}{{3!}} + \frac{{N(N – 1)\,(N – 2)(N – 3)(N – 4)}}{{5!}} + ….} \right]$
$ = \frac{1}{N}\left\{ {{\,^N}{C_1} + {\,^N}{C_3} + {\,^N}{C_5} + ….} \right\}$
$ = \frac{1}{N}\left\{ {{2^{N – 1}}} \right\} = \frac{{{2^n}}}{{n + 1}}$ $\{ N = n + 1\} $
Trick : Put $n=1$, then ${S_1} = \frac{{^1{C_0}}}{1} = \frac{1}{1} = 1$
At $n=2$, ${S_2} = \frac{{^2{C_0}}}{1} + \frac{{^2{C_2}}}{3} = 1 + \frac{1}{3} = \frac{4}{3}$
Also $(c)$ $ \Rightarrow \,\,\,{S_1} = 1,{S_2} = \frac{4}{3}$
