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 $(S_{n-1},0)$ and radius $a_n$ and $D_n$ denote the circle with center $(S_{n-1},S_{n-1})$ and radius $a_n$.

Consider $M$ with $r=\frac{(2^{199}-1)\sqrt{2}}{2^{198}}$. The number of all those circles $D_n$ that are inside $M$ is

• A. $198$
• B. $199$
• C. $200$
• D. $201$

Answer 1

The correct option is B $199$

$\therefore r=\frac{(2^{199}-1)\sqrt{2}}{2^{198}}$
Now,
$\sqrt{2}{S}_{n-1}+a_n<\left(\frac{2^{199}-1}{2^{198}}\right)\sqrt{2}\Rightarrow 2\sqrt{2}(1-\frac{1}{2^{n-1}})+\frac{1}{2^{n-1}}<\left(\frac{2^{199}-1}{2^{198}}\right)\Rightarrow 2\sqrt{2}-\frac{\sqrt{2}}{2^{n-2}}+\frac{1}{2^{n-1}}<2\sqrt{2}-\frac{\sqrt{2}}{2^{198}}\Rightarrow \frac{1}{2^{n-2}}(\frac{1}{2}-\sqrt{2})<-\frac{\sqrt{2}}{2^{198}}\Rightarrow \frac{2\sqrt{2}-1}{2\cdot 2^{n-2}}>\frac{\sqrt{2}}{2^{198}}\Rightarrow 2^{n-1}<(2-\frac{1}{\sqrt{2}})2^{198}\therefore n\le 199$
$\therefore$ Number of circles $=199$