Difference between revisions of "1950 AHSME Problems/Problem 43"

Latest revision as of 15:14, 9 May 2015

Problem

The sum to infinity of $\frac{1}{7}+\frac {2}{7^2}+\frac{1}{7^3}+\frac{2}{7^4}+\cdots$ is:

$\textbf{(A)}\ \frac{1}{5} \qquad \textbf{(B)}\ \dfrac{1}{24} \qquad \textbf{(C)}\ \dfrac{5}{48} \qquad \textbf{(D)}\ \dfrac{1}{16} \qquad \textbf{(E)}\ \text{None of these}$

Solution

Note that this is $\frac{1}{7}(1+\frac{1}{49}+\frac{1}{49^2}+...)+\frac{2}{49}(1+\frac{1}{49}+...)=\frac{9}{49}(1+\frac{1}{49}+...)$ . Using the formula for a geometric series, we find that this is $\frac{9}{49}(\frac{1}{1-\frac{1}{49}})=\frac{9}{49}(\frac{1}{\frac{48}{49}})=\frac{9}{49}(\frac{49}{48})=\frac{9}{48}=\frac{3}{16} \Rightarrow \mathrm{(E)}$

1950 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 42	Followed by Problem 44
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All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 1: / Line 1: @@
 ==Problem==
-The sum to infinity of <math> \frac{1}{7}\plus{}\frac {2}{7^2}\plus{}\frac{1}{7^3}\plus{}\frac{2}{7^4}\plus{}...</math> is:
+The sum to infinity of <math> \frac{1}{7}+\frac {2}{7^2}+\frac{1}{7^3}+\frac{2}{7^4}+\cdots</math> is:
 <math>\textbf{(A)}\ \frac{1}{5} \qquad

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Difference between revisions of "1950 AHSME Problems/Problem 43"

Latest revision as of 15:14, 9 May 2015

Problem

Solution

See Also