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 1

The correct option is C

$\frac{2^n}{n+1}$

Putting the values of $C_0,C_2,C_4$........., we get

= 1 + $\frac{n(n-1)}{3.2!}$ + $\frac{n(n-1)(n-2)(n-3)}{5.4!}$ + .........

= $\frac{1}{n+1}$[(n+1) + $\frac{(n+1)n(n-1)}{3!}$ + $\frac{(n+1)n(n-1)(n-2)(n-3)}{5!}$ + ..........]

Put n+1 = N

= $\frac{1}{N}$[N + $\frac{N(N-1)(N-2)}{3!}$ + $\frac{N(N-1)(N-2)(N-3)(N-4)}{5!}$ + .........]

= $\frac{1}{N}${${}^N{C}_1$ + ${}^N{C}_3$ + ${}^N{C}_5$ +..........}

= $\frac{1}{N}${$2^{N-1}$} = $\frac{2^n}{n+1}$ { ∵ 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) ⇒ $S_1$ = 1, $S_2$ = $\frac{4}{3}$