Question
$$\frac{{{C_0}}}{1} + \frac{{{C_2}}}{3} + \frac{{{C_4}}}{5} + \frac{{{C_6}}}{7} + ..... = $$
A.
$$\frac{{{2^{n + 1}}}}{{n + 1}}$$
B.
$$\frac{{{2^{n + 1}} - 1}}{{n + 1}}$$
C.
$$\frac{{{2^{n}}}}{{n + 1}}$$
D.
None of these
Answer :
$$\frac{{{2^{n}}}}{{n + 1}}$$
Solution :
Putting the value of $${C_0},{C_2},{C_4}.....,$$ we get
$$\eqalign{
& = 1 + \frac{{n\left( {n - 1} \right)}}{{3 \cdot 2!}} + \frac{{n\left( {n - 1} \right)\left( {n - 2} \right)\left( {n - 3} \right)}}{{5 \cdot 4!}} + ..... \cr
& = \frac{1}{{n + 1}}\left[ {\left( {n + 1} \right) + \frac{{\left( {n + 1} \right)n\left( {n - 1} \right)}}{{3!}} + \frac{{\left( {n + 1} \right)n\left( {n - 1} \right)\left( {n - 2} \right)\left( {n - 3} \right)}}{{5!}} + .....} \right] \cr
& {\text{Put, }}n + 1 = N \cr
& = \frac{1}{N}\left[ {N + \frac{{N\left( {N - 1} \right)\left( {N - 2} \right)}}{{3!}} + \frac{{N\left( {N - 1} \right)\left( {N - 2} \right)\left( {N - 3} \right)\left( {N - 4} \right)}}{{5!}} + .....} \right] \cr
& = \frac{1}{N}\left\{ {^N{C_1} + {\,^N}{C_3} + {\,^N}{C_5} + .....} \right\} \cr
& = \frac{1}{N}\left\{ {{2^{N - 1}}} \right\} = \frac{{{2^n}}}{{n + 1}}\,\,\,\,\left\{ {\because N = n + 1} \right\} \cr} $$