Difference between revisions of "2009 AMC 12A Problems/Problem 15"
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<cmath>=-\frac{1}{2}+\frac{(n+1)i^n}{2}-\frac{ni^{n+1}}{2}</cmath> | <cmath>=-\frac{1}{2}+\frac{(n+1)i^n}{2}-\frac{ni^{n+1}}{2}</cmath> | ||
− | Since the imaginary and the real part must be positive, we know that <math>i^{n+1}=-1</math> or <math>i^{n+1}=-i</math>. By the same line of reason, <math>i^{n}=1,i</math>. This only works for <math>n\equiv 1 | + | Since the imaginary and the real part must be positive, we know that <math>i^{n+1}=-1</math> or <math>i^{n+1}=-i</math>. By the same line of reason, <math>i^{n}=1,i</math>. This only works for <math>n\equiv 1 \mod 4</math>. Therefore, we have: |
<cmath>\frac{-1+n}{2}+\frac{(n+1)i}{2}=48+49i</cmath> | <cmath>\frac{-1+n}{2}+\frac{(n+1)i}{2}=48+49i</cmath> | ||
− | Solving either the real or imaginary part gives <math>\boxed{\mathbf{(D)}97}</math> | + | Solving either the real or imaginary part gives <math>\boxed{\mathbf{(D) }97}</math> |
==Video Solution== | ==Video Solution== |
Latest revision as of 16:34, 3 December 2020
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.
Solution 3 (Fast)
Some may know the equation:
(For those curious, this comes from differentiating the equation for finite geometric sums)
Using this equation, we have
Since the imaginary and the real part must be positive, we know that or . By the same line of reason, . This only works for . Therefore, we have:
Solving either the real or imaginary part gives
Video Solution
~savannahsolver
See Also
2009 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 14 |
Followed by Problem 16 |
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 | |
All AMC 12 Problems and Solutions |
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