2011 AIME II Problems/Problem 5
Problem
The sum of the first terms of a geometric sequence is . The sum of the first terms is . Find the sum of the first 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 .
Solution 3
The sum of the first 2011 terms of the sequence is expressible as .... until . The sum of the 2011 terms following the first 2011 is expressible as .... until . Notice that the latter sum of terms can be expressed as . We also know that the latter sum of terms can be obtained by subtracting 200 from 180, which then means that . The terms from 4023 to 6033 can be expressed as , which is equivalent to . Adding 380 and 162 gives the answer of .
Video Solution
https://www.youtube.com/watch?v=rpYphKOIKRs&t=186s ~anellipticcurveoverq
See also
2011 AIME II (Problems • Answer Key • Resources) | ||
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Followed by Problem 6 | |
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