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

(Created page with "==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: <math>\textbf{(A)}\ \frac{1}{5} \qquad \...")
 
(→‎Problem)
 
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==Problem==
 
==Problem==
  
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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:
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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
 
<math>\textbf{(A)}\ \frac{1}{5} \qquad
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\textbf{(D)}\ \dfrac{1}{16} \qquad
 
\textbf{(D)}\ \dfrac{1}{16} \qquad
 
\textbf{(E)}\ \text{None of these}</math>
 
\textbf{(E)}\ \text{None of these}</math>
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 +
==Solution==
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 +
Note that this is <math>\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}+...)</math>. Using the formula for a geometric series, we find that this is <math>\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)}</math>
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==See Also==
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{{AHSME 50p box|year=1950|num-b=42|num-a=44}}
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[[Category:Introductory Algebra Problems]]
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{{MAA Notice}}

Latest revision as of 16:14, 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 $\frac{9}{49}(\frac{1}{1-\frac{1}{49}})=\frac{9}{49}(\frac{1}{\frac{48}{49}})=\frac{9}{49}(\frac{49}{48})=\frac{9}{48}=\frac{3}{16} \Rightarrow \mathrm{(E)}$

See Also

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

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