Question

Prove that $\frac{(2n!)}{n!}=2^n\left(\mathrm{1.3.5....}\right(2n-1\left)\right)$.

Answer 1

$(2n!)=2n(2n-1)(2n-2)....4.3.2.1$
Taking 2 common from alternative even terms,we get
$(2n!)=(\mathrm{2.2.....}n$ times) $\left[n\right(2n-1\left)\right(n-1)....2.3.1.1]$
$(2n!)=2^n\left[\right(2n-1\left)\right(2n-3)...3.1]\left[n\right(n-1)...2.1]$
$(2n!)=2^n\left[\right(2n-1\left)\right(2n-3)...3.1]n!$
$\frac{(2n!)}{n!}=2^n\left(\mathrm{1.3.5.7...}\right(2n-1\left)\right]$
Hence proved.