1959 AMC 12 Problem 49

Question

For the infinite series

\begin{aligned} &1-\dfrac12-\dfrac14+\dfrac18\\ &{}-\dfrac1{16}-\dfrac1{32}+\dfrac1{64}\\ &{}-\dfrac1{128}-\cdots, \end{aligned}

let

S

be the (limiting) sum. Then

S

equals:

0

\dfrac27

\dfrac67

\dfrac9{32}

\dfrac{27}{32}

Group the terms in consecutive blocks of three

Each block is

\frac{1}{8}

times the preceding block

Group the series as

\begin{aligned} S={}&\left(1-\frac12-\frac14\right)\\ &+\left(\frac18-\frac1{16}-\frac1{32}\right) +\cdots. \end{aligned}

The first block is

\frac{1}{4},

and successive blocks form a geometric series with ratio

\frac{1}{8}.

Hence

S=\frac{\frac{1}{4}}{1-\frac{1}{8}}=\frac27.

Therefore, the correct answer is B.

Problem 49 in other years