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!

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
$$
\frac{\pi}{4}\left(\frac{\pi^2}{3}-3\right)=\frac{\pi^3}{12}-\frac{3\pi}{4}
$$
