How do I find the total area of total area of the red circles C_n,n=1,2,3 (Calculus II) 
I am very confused with this problem. I have been working on it for about an hour and still nothing. Help would be greatly appreciated!
 A: One possible solution for this uses Decartes' Circle Theorem, which states that for four mutually externally tangent circles with radii $a,b,c,d$, then the following equality holds:
$$
2\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{1}{d^2}\right)=\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\right)^2
$$
Let $r_n$ be the radius of circle $n$. Use the distance formula between the center of $C_1$ and the center of one of the big circles to obtain
$$
(1-r_1)^2+1^2=(1+r_1)^2\Rightarrow r_1=\frac{1}{4}
$$
We claim by induction that $r_n=\frac{1}{2n(n+1)}$. The base case is trivial, and given $r_n$ is decreasing, it suffices to show that
$$
2(4(n-1)^2n^2+4n^2(n+1)^2+2)=(2(n-1)n+2n(n+1)+2)^2
$$
Or equivalently that
$$
(4n^2(n^2+1)+1)=(2n^2+1)^2
$$
which is true.
Now we want to find
$$
\frac{\pi}{4}\left(\displaystyle\sum_{n\ge 1} \frac{1}{n^2(n+1)^2}\right)
$$
But since
$$
\frac{1}{n^2(n+1)^2} = \frac{1}{n^2}+\frac{1}{(n+1)^2}+2\left(\frac{1}{n+1}-\frac{1}{n}\right)
$$
The answer is
$$
\frac{\pi}{4}\left(\frac{\pi^2}{3}-3\right)=\frac{\pi^3}{12}-\frac{3\pi}{4}
$$
A: you can obtain r1 from this triangle (sorry I don't have the tools for drawing it):
-Join the center of D with the center of C1, its length is (1 + r1).
-Draw a parallel line to T from C1, this will be a leg of our rectangle triangle and this side has length 1.
-The remaining side is (1 - r1)
And applying Pythagoras you can obtain the length of r1, which, if my calculus are correct, is 1/4
For the remaining circles it is just doing the same using the information obtained from the previous circles, that is, for r2 you will need r1, etc.
A: Let the radius of circle $C_i$ be $r_i$ and its centre be denoted by $\overline{C_i}$.  
From the right triangle $\overline{C}P\overline{C_1}$, we have $1+(1-r_1)^2 = (1+r_1)^2 \implies r_1 = \frac14$.
For $r_2$, look at the right triangle $\overline{C}P\overline{C_2}$.  By Pythagoras, we have $1+(1-\frac12-r_2)^2 = (1+r_2)^2 \implies r = \frac1{12}$
Similarly for $r_3$, look at the right triangle $\overline{C}P\overline{C_3}$.  By Pythagoras, we have $1+(1-\frac12-\frac16-r_3)^2 = (1+r_3)^2 \implies r = \frac1{24}$
In general for $r_i$, look at the right triangle $\overline{C}P\overline{C_i}$.  By Pythagoras, we have 
$$1+(1-(2r_1+2r_2+...+2r_{i-1}+r_i))^2 = (1+r_i)^2$$
You can show that $r_i = \dfrac1{2i(i+1)}$, and you have a tough series to sum to get the area - which if I calculated right is $\frac{\pi}{12} \left(\pi^2-9\right)$...
