Explanation

(c)
Putting the values of $C_{0},\ C_{2},\ C_{4}$…., we get
$1 + \frac{n(n - 1)}{3 \bullet 2!} + \frac{n(n - 1)(n - 2)(n - 3)}{5 \bullet 4!} + \ldots$
$= \frac{1}{n + 1}\lbrack(n + 1) + \frac{(n + 1)n(n - 1)}{3!}\$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ + \frac{(n + 1)n(n - 1)(n - 2)(n - 3)}{5!} + \ldots\ \rbrack$
$Put\ n + 1 = N$
$= \frac{1}{N}\lbrack N + \frac{N(N - 1)(N - 2)}{3!}\ \rbrack$
$\ + \frac{N(N - 1)(N - 2)(N - 3)(N - 4)}{5!} + \ldots$
$= \frac{1}{N}\left\lbrack \ ^{N}C_{1} + \ ^{N}C_{3} + \ ^{N}C_{5} + \ldots \right\rbrack$
$= \frac{1}{N}\left\lbrack 2^{N - 1} \right\rbrack = \frac{2^{n}}{n + 1}\ \lbrack\because N = n + 1\rbrack$