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. |
| + | <cmath>1 + \frac{1}{4}\sqrt{2} + \frac{1}{4} + \frac{1}{16}\sqrt{2} + \frac{1}{16} + ...</cmath> | ||
| + | <cmath>=</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> | ||
| + | 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> | ||
== See also == | == See also == | ||
Revision as of 19:11, 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/c/b/a/cbac2ec7ad58e5c966aeec70c5eefabc2e766447.png)
![\[=\]](http://latex.artofproblemsolving.com/b/7/f/b7f0dbf30cb7ab74a726ffc9b15d33af79a3267c.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
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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
