Difference between revisions of "1950 AHSME Problems/Problem 43"
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==Solution== | ==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> | ||
==See Also== | ==See Also== | ||
Revision as of 08:27, 4 May 2012
Problem
The sum to infinity of $\frac{1}{7}\plus{}\frac {2}{7^2}\plus{}\frac{1}{7^3}\plus{}\frac{2}{7^4}\plus{}...$ (Error compiling LaTeX. Unknown error_msg) 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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