32+54+98+(17)/(16) +(33)/(32)+(65)/(64)-7=

1979 AMC 12 Problem 6

Attempt Problem 6 of the 1979 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 1979 AMC 12 solutions, or check the answer key.

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\frac32+\frac54+\frac98+\frac{17}{16}

+\frac{33}{32}+\frac{65}{64}-7=

-\frac1{64}

-\frac1{16}

0

\frac1{16}

\frac1{64}

Write each fraction as

1

plus a reciprocal power of

2

Sum

\frac12+\frac14+\cdots+\frac1{64}

as a finite geometric series

Each numerator is one more than its denominator, so the expression is

\begin{aligned} 6+\biggl(&\frac12+\frac14+\frac18\\ &+\frac1{16}+\frac1{32}+\frac1{64}\biggr)-7. \end{aligned}

The parenthesized geometric sum is

1-\frac{1}{64}=\frac{63}{64}.

Therefore the value is

6+\frac{63}{64}-7=-\frac{1}{64}.

Therefore, the correct answer is A.