Question

If n is a natural number then ${\left(\frac{n+1}{2}\right)}^n\ge n!$ is true when

• A. n > 1
• B. $n\ge 1$
• C. n > 2
• D. never

Answer 1

Answer: (B)

$n\ge 1$

Given equation,

${\left(\frac{n+1}{2}\right)}^n\ge n!$

For $n=1$,

LHS=${\left(\frac{1+1}{2}\right)}^1=1$

RHS=$1!=1$

For $n=2$,

LHS=${\left(\frac{3}{2}\right)}^2=\frac{9}{4}=2.25$

RHS=$2!=2$

Given condition is true for $n\ge 1$. Hence, it is the answer.