Question

If $U_n={\int }_0^1{x}^n(2-x{)}^{n}dx$ and $V_n={\int }_0^1{x}^n(1-x{)}^{n}dx,n\in N$ and $\frac{U_n}{V_n}=1024,$ then the value of $n$ is equal to

Answer 1

$U_n={\int }_0^1{x}^n(2-x{)}^{n}dx$

Put $x=2t\Rightarrow dx=2dt$
$U_n=2{\int }_0^{1/2}{2}^{2n}.t^n(1-t{)}^{n}dt$

Replacing $t$ by $x$,
$U_n=2{\int }_0^{1/2}{2}^{2n}.x^n(1-x{)}^{n}dt\Rightarrow U_n=2^{2n}.V_n$
$\Rightarrow \frac{U_n}{V_n}=1024=2^{2n}$
$\Rightarrow 2^{10}=2^{2n}$
$\Rightarrow n=5$