Let M = {(x, y) in mathbb{R} times mathbb{R} | vedclass

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(D,B) For the geometric progression $a_n = \frac{1}{2^{n-1}}$,the sum $S_n = \sum_{i=1}^n \frac{1}{2^{i-1}} = \frac{1(1 - (1/2)^n)}{1 - 1/2} = 2(1 - \

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(D,B) For the geometric progression $a_n = \frac{1}{2^{n-1}}$,the sum $S_n = \sum_{i=1}^n \frac{1}{2^{i-1}} = \frac{1(1 - (1/2)^n)}{1 - 1/2} = 2(1 - \frac{1}{2^n}) = 2 - \frac{1}{2^{n-1}}$.$(1)$ The circle $C_n$ has center $(S_{n-1}, 0) = (2 - \frac{1}{2^{n-2}}, 0)$ and radius $a_n = \frac{1}{2^{n-1}}$.$C_n$ is inside $M$ if the distance from the origin to the furthest point on the circle is $\leq r$.The furthest point is at distance $|S_{n-1}| + a_n = (2 - \frac{1}{2^{n-2}}) + \frac{1}{2^{n-1}} = 2 - \frac{2-1}{2^{n-1}} = 2 - \frac{1}{2^{n-1}}$.Given $r = \frac{1025}{513} \approx 1.998$,we need $2 - \frac{1}{2^{n-1}} \leq \frac{1025}{513} \implies \frac{1}{2^{n-1}} \geq 2 - \frac{1025}{513} = \frac{1026-1025}{513} = \frac{1}{513}$.Since $2^9 = 512 Since the circles $C_n$ are centered at $(S_{n-1}, 0)$ with radii $a_n = \frac{1}{2^{n-1}}$,they are disjoint if the distance between centers $|S_n - S_{n-1}| > a_n + a_{n+1}$,which is not the case here as they are tangent. However,the question asks for the maximum number of non-intersecting circles. Since they are tangent,we can pick alternating circles to be disjoint,giving $l = 5$.$3k + 2l = 3(10) + 2(5) = 40$. Correct option is $(D)$.$(2)$ The circle $D_n$ has center $(S_{n-1}, S_{n-1})$ and radius $a_n = \frac{1}{2^{n-1}}$.$D_n$ is inside $M$ if $\sqrt{S_{n-1}^2 + S_{n-1}^2} + a_n \leq r \implies S_{n-1}\sqrt{2} + \frac{1}{2^{n-1}} \leq r$.$(2 - \frac{1}{2^{n-2}})\sqrt{2} + \frac{1}{2^{n-1}} \leq \frac{(2^{199}-1)\sqrt{2}}{2^{198}} = (2 - \frac{1}{2^{198}})\sqrt{2}$.This holds for $n-1 \leq 198$,so $n \leq 199$. Thus the number of circles is $199$. Correct option is $(B)$.

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