Q44 / 90 · KVPY 2014 (SB/SX) · Mathematics

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 $\{1,2,3, \ldots \ldots . .20\}$ for which $1.9 \leq \frac{A_{n}}{A_{n-1}} \leq 2$ is :

  • A
    9
  • B
    10
  • Correct answer
    C
    11
  • D
    12

Solution

$\quad A_{n}=\left({ }^{n} C_{r}\right) \max$ case-I $n=2 k($ even $) \Rightarrow A_{n}={ }^{2 k} C_{k}$ $\Rightarrow n-1=2 k-1$ (odd) $\Rightarrow A_{n-1}={ }^{2 k-1} C_{k-1}$ or ${ }^{2 k-1} C_{k}$ $\Rightarrow 1.9 \leq \frac{\mathrm{A}_{\mathrm{n}}}{\mathrm{A}_{\mathrm{n}-1}} \leq 2$ $\Rightarrow 1.9 \leq \frac{{ }^{2 \mathrm{k}} \mathrm{C}_{\mathrm{k}}}{{ }^{2 \mathrm{k}-1} \mathrm{C}_{\mathrm{k}-1}} \leq 2$ $\Rightarrow 1.9 \leq 2 \leq 2$ which is true for every $n=$ even $n=2,4,6, \ldots \ldots \ldots . .20 \Rightarrow 10$ numbers case-II $n=(2 k-1)$ (odd) $A_{n}={ }^{2 k-1} C_{k}$ or $^{2 k-1} C_{k-1}$ $\Rightarrow A_{n-1}={ }^{n}{k}-2 C_{k-1}$ $\because 1.9 \leq \frac{\mathrm{A}_{\mathrm{n}}}{\mathrm{A}_{\mathrm{n}-1}} \leq 2$ $1.9 \leq \frac{{ }^{2 \mathrm{k}-1} \mathrm{C}_{\mathrm{k}}}{{ }^{2 \mathrm{k}-2} \mathrm{C}_{\mathrm{k}-1}} \leq 2.0$ $1.9 \leq \frac{2 k-1}{k} \leq 2$ $1.9 \mathrm{k} \leq 2 \mathrm{k}-1 \leq 2 \mathrm{k}$ $10 \leq \mathrm{k}$ $\Rightarrow \mathrm{n} \geq 19$ $\begin{aligned} \Rightarrow n &=19 \text { only } \\ \text { total } &=11 \end{aligned}$

Hence, the correct answer is option: 'C'.

More Sets and Relations Questions