Infinite Cones, Finite Surface Area?

First, note that the sum of all the lateral faces of the cones is the same as the area of the unit circle, which is $\pi$ .

Now all we need to do is find the sum of the areas of the circular bases.

We can see that the circumference of the circular bases are $\dfrac{1}{2}$ th, $\dfrac{1}{4}$ th, $\dfrac{1}{8}$ th etc. of the circumference of the unit circle.

Since the circumference of the unit circle is $2\pi$ , the circumferences of the bases are $\pi,\dfrac{\pi}{2}.\dfrac{\pi}{4},\ldots$ .

If the circumference of the base is $n\pi$ , then the radius is $\dfrac{n\pi}{2\pi}=\dfrac{n}{2}$ , and the area is $\dfrac{n^2\pi}{4}$ .

Therefore, the sum of the areas of all the bases is $\dfrac{\pi}{4}+\dfrac{\pi}{16}+\dfrac{\pi}{64}+\cdots = \dfrac{\dfrac{\pi}{4}}{1-\dfrac{1}{4}}=\dfrac{\pi}{3}$ .

Our final answer, is therefore $\pi+\dfrac{\pi}{3}=\dfrac{4}{3}\pi$ , and our answer is $4+3=\boxed{7}$ .