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Problem: Let $${C_n}$$ be an infinite sequence of circles lying in the positive quadrant of the $$XY$$-plane, with strictly decreasing radii and satisfying the following conditions. Each $$C_n$$ touches both the $$X$$-axis and the $$Y$$-axis. Further, for all $$n\geq 1$$, the circle $$C_{n+1}$$ touches the circle $$C_n$$ externally. If $$C_1$$ has radius $$10\: cm$$, then show that the sum of the areas of all these circles is $$\frac{25\pi}{3\sqrt{2}-4} \: cm^2$$.

Solution: Consider the following diagram where the Green line segment is $$R_n$$, the radius of the $$n^{th}$$ circle, and the Yellow line segment is $$R_{n+1}$$.

As we are told about the symmetricity of the figure in the problem we can say that:

$$\sqrt{2}R_{n+1} + R_{n+1} + R_n = \sqrt{2} R_n$$

$$=> R_{n+1}(\sqrt{2}+1)=R_n(\sqrt{2}-1)$$

$$=> R_{n+1}= (3-2\sqrt{2})R_n$$

Let’s say $$=> R_{n+1}= \alpha.R_n$$.

Now the total sum of the areas of the circles is:

$$(\pi R_1^2 + \pi R_2^2 + \cdots ) = \pi (R_1^2 + R_2^2 + R_3^2 + \cdots )$$

Now as $$R_{n+1}= \alpha.R_n$$, we can say that:

$$\pi (R_1^2 + R_2^2 + R_3^2 + \cdots ) = \pi (R_1^2 + \alpha^2 R_1^2 + \alpha^4 R_1^2 + \cdots ) = \pi \frac{R_1^2}{1-\alpha^2}$$ as $$\alpha^2 < 1$$.

Substituting the value of $$\alpha = 3-2\sqrt{2}$$ and $$R_1 = 10 \: cm$$we have,

Sum = $$\frac{25\pi}{3\sqrt{2}-4} \: cm^2$$.

Hence Proved.