Question

If $P\left(n\right):2^n(n-1)!<n^n,$ then the smallest positive integer value of $n$ for which $P\left(n\right)$ is true, is

• A. $2$
• B. $3$
• C. $4$
• D. $5$

Answer 1

The correct option is B $3$

$P\left(n\right):2^n(n-1)!<n^n$
For $n=2,\text{LHS}=4\text{and RHS}=4$
$\Rightarrow \text{LHS}\nless \text{RHS}\text{for}n=2$

For $n=3,\text{LHS}=16\text{and RHS}=27$
$\Rightarrow \text{LHS}<\text{RHS}\text{for}n=3$
Thus, the least positive integer for which $P\left(n\right)$ is true is $3.$