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Finding an infinite geometric series (sorta)

Source: 1959 AHSME Problem 49

August 14, 2013
AMC

Problem Statement

For the infinite series 1−12−14+18−116−132+164−1128−⋯1-\frac12-\frac14+\frac18-\frac{1}{16}-\frac{1}{32}+\frac{1}{64}-\frac{1}{128}-\cdots let SS be the (limiting) sum. Then SS equals: <spanclass=′latex−bold′>(A)</span> 0<spanclass=′latex−bold′>(B)</span> 27<spanclass=′latex−bold′>(C)</span> 67<spanclass=′latex−bold′>(D)</span> 932<spanclass=′latex−bold′>(E)</span> 2732 <span class='latex-bold'>(A)</span>\ 0\qquad<span class='latex-bold'>(B)</span>\ \frac27\qquad<span class='latex-bold'>(C)</span>\ \frac67\qquad<span class='latex-bold'>(D)</span>\ \frac{9}{32}\qquad<span class='latex-bold'>(E)</span>\ \frac{27}{32}
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