Question

$(2n+1)(2n+3)(2n+5)\dots (4n-1)=$

• A. $\frac{\left(4n\right)!}{2^n\left[\right(2n)!{]}^2}$
• B. $\frac{\left(4n\right)!}{\left[\right(2n)!{]}^2}$
• C. $\frac{\left(4n\right)!n!}{2^n\left[\right(2n)!{]}^2}$
• D. $\frac{\left(4n\right)!n!}{\left[\right(2n)!{]}^2}$

Answer 1

Answer: (C)

$\frac{\left(4n\right)!n!}{2^n\left[\right(2n)!{]}^2}$

$(2n+1)(2n+3)(2n+5)\dots .(4n-l)$
(On multiplying and dividing by $2n!$, also multiply and divide by $(2n+2)(2n+4)...4n$)
$\frac{\left(2n\right)!(2n+1)(2n+2)\dots (4n.-1)\left(4n\right)}{\left(2n\right)!(2n+2)(2n+4)..4n}$
(In the denominator, take factor $2$ common from the terms $2n+2,2n+\mathrm{4...}$) to get
$\frac{\left(4n\right)!}{\left(2n\right)!2^n(n+1)(n+2)\dots \left(2n\right)}$

Now, multiplying and dividing by $n!$, we get
$\frac{\left(4n\right)!n!}{\left(2n\right)!2^n.n!(n+1)(n+2)...\left(2n\right)}=\frac{\left(4n\right)!n!}{2^n{\left[\right(2n)!]}^2}$