Question

Let $M=\left\{(x, y) \in R \times R: x^{2}+y^{2} \leq r^{2}\right\}$, where $r>0$. Consider the geometric progression $a_{n}=\frac{1}{2^{n-1}}, n=1,2,3, \ldots .$ Let $S_{0}=0$ and for $n \geq 1$, let $S_{n}$ denote the sum of the first $n$ terms of this progression. For $n \geq 1$, let $C_{n}$ denote the circle with center $\left(S_{n-1}, 0\right)$ and radius $a_{n}$, and $D_{n}$ denote the circle with center $\left(S_{n-1}, S_{n-1}\right)$ and radius $a_{n}$.
Consider $M$ with $r=\frac{\left(2^{199}-1\right) \sqrt{2}}{2^{198}}$. The number of all those circles $D_{n}$ that are inside $M$ is

• A. 198
• B. 199
• C. 200
• D. 201

Answer 1

Answer: (B)

$\sqrt{2} S_{n-1}+a_{n} < r$
$\Rightarrow \sqrt{2}\left(2\left(1-\frac{1}{2^{n-1}}\right)\right)+\frac{1}{2^{n-1}} < \left(\frac{2^{199}-1}{2^{198}}\right) \sqrt{2}$
$\Rightarrow \sqrt{2}\left(1-\frac{1}{2^{n-1}}\right)+\frac{1}{2^{n}} < \left(1-\frac{1}{2^{199}}\right) \sqrt{2}$
$\Rightarrow \frac{1}{(\sqrt{2})^{2 n}}-\frac{1}{(\sqrt{2})^{2 n-3}} < \frac{-1}{(\sqrt{2})^{397}}$
$\Rightarrow \frac{2 \sqrt{2}-1}{(\sqrt{2})^{2 n}}>\frac{1}{(\sqrt{2})^{397}}$
$\Rightarrow (\sqrt{2})^{2 n-397} < 2 \sqrt{2}-1$
$\Rightarrow 2 n-397 \leq 1 \Rightarrow n \leq 199$