1959 AMC 12 Problem 49

Attempt Problem 49 of the 1959 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1959 AMC 12 solutions, or check the answer key.

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49.

For the infinite series 11214+18116132+1641128, \begin{aligned} &1-\dfrac12-\dfrac14+\dfrac18\\ &{}-\dfrac1{16}-\dfrac1{32}+\dfrac1{64}\\ &{}-\dfrac1{128}-\cdots, \end{aligned} let SS be the (limiting) sum. Then SS equals:

00

27\dfrac27

67\dfrac67

932\dfrac9{32}

2732\dfrac{27}{32}

Answer: B
Concepts:geometric sequencepairing and groupingsummation
Difficulty rating: 1590
Small Hint:

Group the terms in consecutive blocks of three

Big Hint:

Each block is 18\frac{1}{8} times the preceding block

Solution:

Group the series as S=(11214)+(18116132)+. \begin{aligned} S={}&\left(1-\frac12-\frac14\right)\\ &+\left(\frac18-\frac1{16}-\frac1{32}\right) +\cdots. \end{aligned} The first block is 14,\frac{1}{4}, and successive blocks form a geometric series with ratio 18.\frac{1}{8}. Hence S=14118=27. S=\frac{\frac{1}{4}}{1-\frac{1}{8}}=\frac27.

Therefore, the correct answer is B.

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Problem 49 in Other Years

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