Difference between revisions of "1952 AHSME Problems/Problem 50"
m (Corrected AHSME box format) |
m (→Problem) |
||
| (3 intermediate revisions by one other user not shown) | |||
| Line 2: | Line 2: | ||
A line initially 1 inch long grows according to the following law, where the first term is the initial length. | A line initially 1 inch long grows according to the following law, where the first term is the initial length. | ||
| − | < | + | <math>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</math> |
If the growth process continues forever, the limit of the length of the line is: | If the growth process continues forever, the limit of the length of the line is: | ||
| Line 13: | Line 13: | ||
== 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>(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> | ||
| + | <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)}</math> | ||
== See also == | == See also == | ||
Latest revision as of 11:51, 9 May 2024
Problem
A line initially 1 inch long grows according to the following law, where the first term is the initial length.
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 
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. 
