1950 AHSME Problems/Problem 43

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
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50
All AHSME Problems and Solutions

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

Page

Toolbox

Search

1950 AHSME Problems/Problem 43

Problem

Solution

See Also