2009 AMC 12A Problems/Problem 15
Contents
Problem
For what value of is ?
Note: here .
Solution 1
We know that cycles every powers so we group the sum in s.
We can postulate that every group of is equal to . For 24 groups we thus, get as our sum. We know the solution must lie near The next term is the th term. This term is equal to (first in a group of so ) and our sum is now so is our answer
Solution 2
Obviously, even powers of are real and odd powers of are imaginary. Hence the real part of the sum is , and the imaginary part is .
Let's take a look at the real part first. We have , hence the real part simplifies to . If there were an odd number of terms, we could pair them as follows: , hence the result would be negative. As we need the real part to be , we must have an even number of terms. If we have an even number of terms, we can pair them as . Each parenthesis is equal to , thus there are of them, and the last value used is . This happens for and . As is not present as an option, we may conclude that the answer is .
In a complete solution, we should now verify which of and will give us the correct imaginary part.
We can rewrite the imaginary part as follows: . We need to obtain . Once again we can repeat the same reasoning: If the number of terms were even, the left hand side would be negative, thus the number of terms is odd. The left hand side can then be rewritten as . We need parentheses, therefore the last value used is . This happens when or , and we are done.
See Also
2009 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 14 |
Followed by Problem 16 |
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All AMC 12 Problems and Solutions |
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