1953 AHSME Problems/Problem 27

The radius of the first circle is $1$ inch, that of the second $\frac{1}{2}$ inch, that of the third $\frac{1}{4}$ inch and so on indefinitely. The sum of the areas of the circles is:

$\textbf{(A)}\ \frac{3\pi}{4} \qquad \textbf{(B)}\ 1.3\pi \qquad \textbf{(C)}\ 2\pi \qquad \textbf{(D)}\ \frac{4\pi}{3}\qquad \textbf{(E)}\ \text{none of these}$

Note the areas of these triangles is $1\pi$ , $\frac{\pi}{4}$ , $\frac{\pi}{16}, \dots$ . The sum of these areas will thus be $\pi\left(1+\frac{1}{4}+\frac{1}{16}+\dots\right)$ . We use the formula for an infinite geometric series to get the sum of the areas will be $\pi\left(\frac{1}{1-\frac{1}{4}}\right)$ , or $\frac{4\pi}{3}$ . $\boxed{D}$

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1953 AHSME Problems/Problem 27