Question

Show that $|\underset{–}{1}|\underset{–}{3}|\underset{–}{5}....|\underset{–}{2n-1}>{\left(|\underset{–}{n}\right)}^n$.

Answer 1

$\frac{\underset{–}{|r}\underset{–}{|2n-r}}{\underset{–}{|n}}>\underset{–}{|n}$

$\frac{\underset{–}{|1}\underset{–}{|2n-1}}{\underset{–}{|n}}>\underset{–}{|n}$

$\frac{\underset{–}{|2}\underset{–}{|2n-2}}{\underset{–}{|n}}>\underset{–}{|n}$

$\frac{\underset{–}{|3}\underset{–}{|2n-3}}{\underset{–}{|n}}>\underset{–}{|n}$

.

.

.

$\frac{\underset{–}{|2n-1}\underset{–}{|1}}{\underset{–}{|n}}>\underset{–}{|n}$

Multiplying the above equation we get .

$\frac{{(\underset{–}{|1}\underset{–}{|2}\underset{–}{|3}...\underset{–}{|2n-1})}^2}{{\underset{–}{|n}}^n}>{\underset{–}{|n}}^n$

${(\underset{–}{|1}\underset{–}{|2}\underset{–}{|3}...\underset{–}{|2n-1})}^2>{\underset{–}{|n}}^{2n}$

$(\underset{–}{|1}\underset{–}{|2}\underset{–}{|3}...\underset{–}{|2n-1})>{\underset{–}{|n}}^n$