For each positive integer n, let A_n= ( l n r ) 0 r n . Then, the… - Vidyadip

MCQ

For each positive integer $n$, let $A_n=\max \left\{\left(\begin{array}{l}n \\ r\end{array}\right) \mid 0 \leq r \leq n\right\}$. Then, the number of elements $n$ is $\{1,2, \ldots, 20\}$ for which $1.9 \leq \frac{A_n}{A_{n-1}} \leq 2$ is

A
$9$
B
$10$
✓
$11$
D
$12$

✓

Answer

Correct option: C.

$11$

We have,
$A_n =\max \left\{{ }^n C_r \mid 0 \leq r \leq n\right\}$
$n \in\{1,2,3, \ldots, 20\}$
Case $I$ When $n$ is even
$A_n={ }^n C_{n / 2}$
$\therefore \frac{A_n}{A_{n-1}}=\frac{{ }^n C_{n ' 2}}{{ }^{n-1} C_{\frac{n-1-1}{}}^2}=2$
So for all $n$ even given relation is true.
Case $II$ When $n$ is odd
$A_n={ }^n C_{\frac{n-1}{}}^2$
$\therefore \frac{A_n}{A_{n-1}} =\frac{{ }^n C_{n-1}}{{ }^{n-1} C_{n-1}}=\frac{2 n}{n+1}$
$19 \leq \frac{2 n}{n+1} \leq n$ if $n=19$
$\therefore$ Total number of elements are $10$ even number and $19=10+1=11$

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