Consider $M$ with $r=\frac{1025}{513}$. Let $k$ be the number of all those circles $C_{n}$ that are inside $M$. Let $1$ be the maximum possible number of circles among these $k$ circles such that no two circles intersect Then
Step 1: Understand the given expression for radius $r$
We are given the radius $r$ of the circle $M$ as:
$$ r = \frac{1025}{513}. $$
We are tasked with finding the number of all circles $C_n$ that are inside $M$ and determining a specific value based on the given conditions.
Step 2: Consider the number of circles $k$ inside $M$
Let $k$ be the number of circles $C_n$ inside $M$ such that no two circles intersect. We are also given that the maximum possible number of circles, among these $k$ circles, such that no two circles intersect is 1.
Step 3: Analyze the general form of the radius and the sum of areas
We are given that:
$$ a_n = \frac{1}{2^{n-1}} $$
and
$$ S_n = 2 \left( 1 - \frac{1}{2^n} \right). $$
These represent the areas of the circles $C_n$ inside $M$. To ensure the circles fit inside $M$, the sum of areas for circles $C_{n-1}$ and $C_n$ should be less than the area of $M$.
Step 4: Apply the condition for non-intersecting circles
We are given the condition:
$$ S_{n-1} + a_n < \frac{1025}{513}. $$
Substituting for $S_n$ and $a_n$, we get:
$$ 2 \left( 1 - \frac{1}{2^n} \right) < \frac{1025}{513}. $$
Simplifying further, we get:
$$ 1 - \frac{1}{2^n} < \frac{1025}{1026}, $$
which simplifies to:
$$ \frac{1}{2^n} > \frac{1}{1026}. $$
Thus, we have:
$$ 2^n < 1026. $$
Taking the logarithm, we get:
$$ n ≤ 10. $$
Thus, the maximum number of circles $n$ is 10.
Step 5: Maximum number of non-intersecting circles
It’s evident that alternate circles, namely $C_1, C_3, C_5, C_7, C_9$, do not intersect each other, and similarly, $C_2, C_4, C_6, C_8, C_{10}$ do not intersect each other. Therefore, a maximum of 5 sets of circles do not intersect each other.
Thus, the value of $l = 5$.
Step 6: Calculate the final result
We are asked to find the value of $3K + 2l$. Substituting the known values, we get:
$$ 3K + 2l = 3 × 10 + 2 × 5 = 30 + 10 = 40. $$
Step 7: Conclusion
Thus, the correct answer is Option (D), which is $\boxed{40}$.
If \(\sum\)\(_{r=1}^n T_r\) = \(\frac{(2n-1)(2n+1)(2n+3)(2n+5)}{64}\) , then \( \lim_{n \to \infty} \sum_{r=1}^n \frac{1}{T_r} \) is equal to :
Figure 1 shows the configuration of main scale and Vernier scale before measurement. Fig. 2 shows the configuration corresponding to the measurement of diameter $ D $ of a tube. The measured value of $ D $ is: