Difference between revisions of "1952 AHSME Problems/Problem 50"
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== Solution == | == Solution == | ||
We can rewrite our sum as the sum of two infinite geometric sequences. | We can rewrite our sum as the sum of two infinite geometric sequences. | ||
| − | <cmath>1 + \frac{1}{4}\sqrt{2} + \frac{1}{4} + \frac{1}{16}\sqrt{2} + \frac{1}{16} + ... | + | <cmath>1 + \frac{1}{4}\sqrt{2} + \frac{1}{4} + \frac{1}{16}\sqrt{2} + \frac{1}{16} + ... = </cmath> |
| − | |||
<cmath>(1 + \frac{1}{4} + \frac{1}{16} + \frac{1}{64} + ...) + (\frac{1}{4}\sqrt{2} + \frac{1}{16}\sqrt{2} + \frac{1}{64}\sqrt{2} + ...)</cmath> | <cmath>(1 + \frac{1}{4} + \frac{1}{16} + \frac{1}{64} + ...) + (\frac{1}{4}\sqrt{2} + \frac{1}{16}\sqrt{2} + \frac{1}{64}\sqrt{2} + ...)</cmath> | ||
<cmath>(1 + \frac{1}{4} + \frac{1}{16} + \frac{1}{64} + ...) + \sqrt{2}(\frac{1}{4} + \frac{1}{16} + \frac{1}{64} + ...)</cmath> | <cmath>(1 + \frac{1}{4} + \frac{1}{16} + \frac{1}{64} + ...) + \sqrt{2}(\frac{1}{4} + \frac{1}{16} + \frac{1}{64} + ...)</cmath> | ||
We now take the sum of each of the infinite geometric sequences separately | We now take the sum of each of the infinite geometric sequences separately | ||
| − | <cmath>(1 + \frac{1}{4} + \frac{1}{16} + \frac{1}{64} + ...) + \sqrt{2}(\frac{1}{4} + \frac{1}{16} + \frac{1}{64} + ...)</cmath> | + | <cmath>(1 + \frac{1}{4} + \frac{1}{16} + \frac{1}{64} + ...) + \sqrt{2}(\frac{1}{4} + \frac{1}{16} + \frac{1}{64} + ...) = </cmath> |
| + | <cmath>\frac{1}{1 - \frac{1}{4}} + \sqrt{2}(\frac{\frac{1}{4}}{1 - \frac{1}{4}})</cmath> | ||
| + | <cmath>\frac{4}{3} + \frac{\sqrt{2}}{3}</cmath> | ||
| + | <cmath>\frac{1}{3}(4 + \sqrt{2})</cmath> | ||
| + | |||
| + | Therefore, the answer is <math>\fbox{(D) \frac{1}{3}(4 + \sqrt{2})}</math> | ||
== See also == | == See also == | ||
Revision as of 19:16, 22 December 2015
Problem
A line initially 1 inch long grows according to the following law, where the first term is the initial length.
\[1+\frac{1}{4}\sqrt{2}+\frac{1}{4}+\frac{1}{16}\sqrt{2}+\frac{1}{16}+\frac{1}{64}\sqrt{2}+\frac{1}{64}+\cdots\] (Error making remote request. Unexpected URL sent back)
If the growth process continues forever, the limit of the length of the line is:
Solution
We can rewrite our sum as the sum of two infinite geometric sequences.
![\[1 + \frac{1}{4}\sqrt{2} + \frac{1}{4} + \frac{1}{16}\sqrt{2} + \frac{1}{16} + ... =\]](http://latex.artofproblemsolving.com/1/7/c/17c0e6087923a043218216aba5ec0f7adf86895f.png)
![\[(1 + \frac{1}{4} + \frac{1}{16} + \frac{1}{64} + ...) + (\frac{1}{4}\sqrt{2} + \frac{1}{16}\sqrt{2} + \frac{1}{64}\sqrt{2} + ...)\]](http://latex.artofproblemsolving.com/3/9/b/39ba564284fead5f022dbc0b50411414dcd0b368.png)
![\[(1 + \frac{1}{4} + \frac{1}{16} + \frac{1}{64} + ...) + \sqrt{2}(\frac{1}{4} + \frac{1}{16} + \frac{1}{64} + ...)\]](http://latex.artofproblemsolving.com/f/5/3/f5358e5bbcff092cee0ea98f5dbf2de0562d2f7b.png)
We now take the sum of each of the infinite geometric sequences separately
![\[(1 + \frac{1}{4} + \frac{1}{16} + \frac{1}{64} + ...) + \sqrt{2}(\frac{1}{4} + \frac{1}{16} + \frac{1}{64} + ...) =\]](http://latex.artofproblemsolving.com/e/5/f/e5f18c8bb7060b3c8df78e9e7512443eb12b6015.png)
![\[\frac{1}{1 - \frac{1}{4}} + \sqrt{2}(\frac{\frac{1}{4}}{1 - \frac{1}{4}})\]](http://latex.artofproblemsolving.com/2/5/d/25dae0e8297aff70f3a3df152ebdfb465753dab0.png)
![\[\frac{4}{3} + \frac{\sqrt{2}}{3}\]](http://latex.artofproblemsolving.com/1/3/d/13d379dbecdc4c01e0227077dd929bc0c47e5538.png)
![\[\frac{1}{3}(4 + \sqrt{2})\]](http://latex.artofproblemsolving.com/c/3/3/c33adc358068870a0e6c88777e56d91eb0c63070.png)
Therefore, the answer is $\fbox{(D) \frac{1}{3}(4 + \sqrt{2})}$ (Error compiling LaTeX. Unknown error_msg)
See also
| 1952 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 49 |
Followed by Last Question | |
| 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. 
