Difference between revisions of "2011 AIME II Problems/Problem 5"
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Thus, <math>200+180+162=542</math>, so the sum of the first <math>6033</math> terms is <math>\boxed{542}</math>. | Thus, <math>200+180+162=542</math>, so the sum of the first <math>6033</math> terms is <math>\boxed{542}</math>. | ||
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+ | ==Solution 2== | ||
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+ | Solution by e_power_pi_times_i | ||
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+ | The sum of the first <math>2011</math> terms can be written as <math>\dfrac{a_1(1-k^{2011})}{1-k}</math>, and the first <math>4022</math> terms can be written as <math>\dfrac{a_1(1-k^{4022})}{1-k}</math>. Dividing these equations, we get <math>\dfrac{1-k^{2011}}{1-k^{4022}} = \dfrac{10}{19}</math>. Noticing that <math>k^{4022}</math> is just the square of <math>k^{2011}</math>, we substitute <math>x = k^{2011}</math>, so <math>\dfrac{1}{x+1} = \dfrac{10}{19}</math>. That means that <math>k^{2011} = \dfrac{9}{10}</math>. Since the sum of the first <math>6033</math> terms can be written as <math>\dfrac{a_1(1-k^{6033})}{1-k}</math>, dividing gives <math>\dfrac{1-k^{2011}}{1-k^{6033}}</math>. Since <math>k^{6033} = \dfrac{729}{1000}</math>, plugging all the values in gives <math>\boxed{542}</math>. | ||
==See also== | ==See also== |
Revision as of 12:50, 1 July 2016
Contents
Problem
The sum of the first 2011 terms of a geometric sequence is 200. The sum of the first 4022 terms is 380. Find the sum of the first 6033 terms.
Solution
Since the sum of the first terms is , and the sum of the fist terms is , the sum of the second terms is . This is decreasing from the first 2011, so the common ratio is less than one.
Because it is a geometric sequence and the sum of the first 2011 terms is , second is , the ratio of the second terms to the first terms is . Following the same pattern, the sum of the third terms is .
Thus, , so the sum of the first terms is .
Solution 2
Solution by e_power_pi_times_i
The sum of the first terms can be written as , and the first terms can be written as . Dividing these equations, we get . Noticing that is just the square of , we substitute , so . That means that . Since the sum of the first terms can be written as , dividing gives . Since , plugging all the values in gives .
See also
2011 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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