Difference between revisions of "1950 AHSME Problems/Problem 43"
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==Problem== | ==Problem== | ||
| − | The sum to infinity of <math> \frac{1}{7} | + | 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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==Solution== | ==Solution== | ||
| − | {{ | + | |
| + | 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> | ||
==See Also== | ==See Also== | ||
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[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 15:14, 9 May 2015
Problem
The sum to infinity of 
is:
Solution
Note that this is 
. Using the formula for a geometric series, we find that this is 
See Also
| 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. 
