Question
The value of $$\frac{{{C_1}}}{2} + \frac{{{C_3}}}{4} + \frac{{{C_5}}}{6} + .....$$ is equal to
A.
$$\frac{{{2^n} + 1}}{{n + 1}}$$
B.
$$\frac{{{2^n} }}{{n + 1}}$$
C.
$$\frac{{{2^n} + 1}}{{n - 1}}$$
D.
$$\frac{{{2^n} - 1}}{{n + 1}}$$
Answer :
$$\frac{{{2^n} - 1}}{{n + 1}}$$
Solution :
$$\eqalign{
& \frac{{{C_1}}}{2} + \frac{{{C_3}}}{4} + \frac{{{C_5}}}{6} + ..... = \frac{n}{2} + \frac{{n\left( {n - 1} \right)\left( {n - 2} \right)}}{{4!}} + \frac{{n\left( {n - 1} \right)\left( {n - 2} \right)\left( {n - 3} \right)\left( {n - 4} \right)}}{{6!}} + ..... \cr
& = \frac{1}{{n + 1}}\left[ {\frac{{\left( {n + 1} \right)n}}{{2!}} + \frac{{\left( {n + 1} \right)n\left( {n - 1} \right)\left( {n - 2} \right)}}{{4!}} + \frac{{\left( {n + 1} \right)n\left( {n - 1} \right)\left( {n - 2} \right)\left( {n - 3} \right)\left( {n - 4} \right)}}{{6!}} + .....} \right] \cr
& = \frac{1}{{n + 1}}\left[ {^{n + 1}{C_2} + {\,^{n + 1}}{C_4} + {\,^{n + 1}}{C_6} + .....} \right] \cr
& = \frac{1}{{n + 1}}\left[ {{2^{n + 1 - 1}} - {\,^{n + 1}}{C_0}} \right] = \frac{{{2^n} - 1}}{{n + 1}} \cr} $$