Question

The positive integer $k$ for which $\frac{(101{)}^{\frac{k}{2}}}{k!}$ is maximum is:

• A. $9$
• B. $10$
• C. $11$
• D. $101$

Answer 1

The correct option is B $10$

Let $f_k=(101{)}^{k/2}/k!$

Let maximum value occurs for $k=n$

Then $f_n>f_{n+1}$

$\Rightarrow (101{)}^{n/2}/n!>(101{)}^{(n+1)/2}/(n+1)!$

$\Rightarrow 1>\left(101{)}^{1/2}/\right(n+1)$

$\Rightarrow n+1>(101{)}^{1/2}$

$\Rightarrow n>(101{)}^{1/2}-1$

Now $(101{)}^{1/2}-1$ is slightly greater than $10$ which means $(101{)}^{1/2}-1$ greater than $9$.

Hence next integer after $9$ is $n=10$