1959 AHSME Problems/Problem 49

Problem 49

For the infinite series $1-\frac12-\frac14+\frac18-\frac{1}{16}-\frac{1}{32}+\frac{1}{64}-\frac{1}{128}-\cdots$ let $S$ be the (limiting) sum. Then $S$ equals:

$\textbf{(A)}\ 0\qquad\textbf{(B)}\ \frac27\qquad\textbf{(C)}\ \frac67\qquad\textbf{(D)}\ \frac{9}{32}\qquad\textbf{(E)}\ \frac{27}{32}$

Solution

Notice that we can arrange the sequence like so: $\begin{align*}1-\frac12-\frac14+\frac18-\frac{1}{16}-\frac{1}{32}+\frac{1}{64}-\frac{1}{128}-\cdots &= 1-\frac{5}{8}-\frac{5}{64}-\frac{5}{512}-\dots \\ &= 1-\left(\sum_{i=1}^{\infty}\frac{5}{8^n}\right) \\ &= 1-\frac{5}{7} \text{ (by the convergence of a geometric series)} \\ &=\frac{2}{7}\end{align*}$ hence our answer is $\fbox{B}$

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1959 AHSME Problems/Problem 49

Problem 49

Solution