Question

Let $M=\left\{(x,y)\in R\times R:x^2+y^2\le r^2\right\}$, where $r>0$. Consider the geometric progression $a_n=\frac{1}{2^{n-1}},n=1,2,3,\dots .$ Let $S_0=0$ and for $n\ge 1$, let $S_n$ denote the sum of the first $n$ terms of this progression. For $n\ge 1$, 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)

$\sqrt{2}{S}_{n-1}+a_n<r$
$\Rightarrow \sqrt{2}\left(2\left(1-\frac{1}{2^{n-1}}\right)\right)+\frac{1}{2^{n-1}}<\left(\frac{2^{199}-1}{2^{198}}\right)\sqrt{2}$
$\Rightarrow \sqrt{2}\left(1-\frac{1}{2^{n-1}}\right)+\frac{1}{2^n}<\left(1-\frac{1}{2^{199}}\right)\sqrt{2}$
$\Rightarrow \frac{1}{(\sqrt{2}{)}^{2n}}-\frac{1}{(\sqrt{2}{)}^{2n-3}}<\frac{-1}{(\sqrt{2}{)}^{397}}$
$\Rightarrow \frac{2\sqrt{2}-1}{(\sqrt{2}{)}^{2n}}>\frac{1}{(\sqrt{2}{)}^{397}}$
$\Rightarrow (\sqrt{2}{)}^{2n-397}<2\sqrt{2}-1$
$\Rightarrow 2n-397\le 1\Rightarrow n\le 199$