Question

The value of $\frac{1}{(2n-1)!0!}+\frac{1}{(2n-3)!2!}+\frac{1}{(2n-5)!4!}+....+\frac{1}{1!(2n-2)!}$ equal to

• A. $2^{2n-1}$
• B. $2^{2n-2}$
• C. $2^{2n-3}$
• D. $\frac{2^{2n-2}}{(2n-1)!}$

Answer 1

The correct option is D $\frac{2^{2n-2}}{(2n-1)!}$

We know that,

${}^n{C}_0{+}^n{C}_1{+}^n{C}_2.........{.}^n{C}_n=2^n$

${}^n{C}_0{+}^n{C}_2{+}^n{C}_4.........{.}^n{C}_{n-1}=$${}^n{C}_1{+}^n{C}_3{+}^n{C}_5.........{.}^n{C}_n=2^{n-1}$

Using the above relation,

${}^{2n-1}{C}_0{+}^{2n-1}{C}_1{+}^{2n-1}{C}_2.........{.}^{2n-1}{C}_n=2^{(2n-1)-1}$

Expanding the coefficients we get the desired result,

$\frac{(2n-1)!}{(2n-1)!0!}+$$\frac{(2n-1)!}{(2n-3)!2!}+$$\frac{(2n-1)!}{(2n-5)!4!}+$$...............+\frac{(2n-1)!}{1!(2n-2!)}=2^{2n-2}$

$\frac{1}{(2n-1)!0!}+$$\frac{1}{(2n-3)!2!}+$$\frac{1}{(2n-5)!4!}+$$...............+\frac{1}{1!(2n-2!)}=\frac{2^{2n-2}}{(2n-1)!}$