Difference between revisions of "1952 AHSME Problems/Problem 50"
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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 14: | Line 14: | ||
== 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)}</math> | ||
== See also == | == See also == |
Latest revision as of 12:51, 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.
We now take the sum of each of the infinite geometric sequences separately
Therefore, the answer is
See also
1952 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 49 |
Followed by Last Question | |
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