Question

Which of the following is equal to $\frac{\mathrm{1.3.5....}(2n-1)}{\mathrm{2.4.6....}\left(2n\right)}$?

• A. $(2n!)÷\left(2^n\right(n!){)}^2$
• B. $(2n!)÷n!$
• C. $(2n-1)÷(n-1)!$
• D. $2^n$

Answer 1

The correct option is A $(2n!)÷\left(2^n\right(n!){)}^2$

$\frac{\mathrm{1.3.5...}(2n-1)}{\mathrm{2.4.6...}\left(2n\right)}⟹\frac{\mathrm{1.2.3.4...}(2n-1).\left(2n\right)}{{\left(\mathrm{2.4.6...}\right(2n\left)\right)}^2}⟹\frac{2n!}{{\left(\right(2.1\left)\right(2.2).(2.3)...(2.n\left)\right)}^2}⟹\frac{2n!}{{(2^{n}n!)}^2}⟹\frac{2n!}{{\left(2^{2n}\right)(n!)}^2}$