Question

For each positive integer $n$, let $A_n=$max$\{\left[\begin{array}{c}n\\ r\end{array}\right];0\le r\le n\}$. Then the number of elements $n$ in $\{1,2,3,.....20\}$ for which $1.9\le \frac{A_n}{A_{n-1}}\le 2$ is$?$

• A. $9$
• B. $10$
• C. $11$
• D. $12$

Answer 1

Answer: (C)

$11$

$\text{Case 1}:$

When $n$ is even, we have $A_n=$ ${}^n{C}_{n/2}$ and $A_{n-1}=$ ${}^{n-1}{C}_{(n-2)/2}$

Thus,$\frac{A_n}{A_{n-1}}=\frac{n!}{\left(n/2\right)!\left(n/2\right)!}\frac{\left(n/2\right)!\left(\right(n-2\left)/2\right)!}{(n-1)!}=\frac{n}{n/2}=2$

Thus, for all even numbers $n$, $1.9\le \frac{A_n}{A_{n-1}}\le 2$

$\text{Case 2}:$

When $n$ is odd, we have $A_n=$ ${}^n{C}_{(n-1)/2}$ and $A_{n-1}=$ ${}^{n-1}{C}_{(n-1)/2}$

Thus,$\frac{A_n}{A_{n-1}}=\frac{n!}{\left(\right(n-1\left)/2\right)!\left(\right(n+1\left)/2\right)!}\frac{\left(\right(n-1\left)/2\right)!(n-1)/2)!}{(n-1)!}=\frac{n}{(n+1)/2}=\frac{2n}{n+1}$

We have $1.9<\frac{2n}{n+1}\le 2⟺n>19$

Only one such $n=20$ is possible.

Total no. of $n$'s in both the cases $=$ $10+1=11$