Difference between revisions of "1964 AHSME Problems/Problem 34"
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<math>\textbf{(D) }\frac{1}{2}[(n+1)(1-i)+2]\qquad \textbf{(E) }\frac{1}{8}(n^2+8-4ni)</math> | <math>\textbf{(D) }\frac{1}{2}[(n+1)(1-i)+2]\qquad \textbf{(E) }\frac{1}{8}(n^2+8-4ni)</math> | ||
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+ | ==Solution== | ||
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+ | The real part is <math>1 + 3i^2 + 5i^4 + ...</math>, which is <math>1 - 3 + 5 - 7 + ...</math>. If <math>n</math> is a multiple of <math>4</math>, then we have an odd number of terms in total: we start with <math>1</math> at <math>n=0</math>, then add two more terms to get <math>1 - 3 + 5</math> at <math>n=4</math>, etc. With each successive addition, we're really adding a total of <math>2</math>, since <math>-3 + 5 = 2</math>, and <math>-7 + 9 = 2</math>, etc. | ||
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+ | At <math>n = 0</math>, the sum is <math>1</math>, and at <math>n=4</math>, the sum is <math>3</math>. Since the sum increases linearly, the real part of the sum is <math>1 + \frac{1}{2}n</math>. | ||
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+ | The imaginary part is <math>\frac{1}{i}(2i + 4i^3 + 6i^5 + ...)</math>, which is <math>2 - 4 + 6 - 8 + ...</math>. This time, we have an even number of terms. We group pairs of terms to get <math>(2-4) + (6-8) + ...</math>, and notice that each pair gives <math>-2</math>. Again, with <math>n=0</math> the imaginary part is <math>0</math>, while with <math>n=4</math> the imaginary part is <math>-2</math>. Since again the sum increases linearly, this means the imaginary part is <math>-\frac{n}{2}</math>. | ||
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+ | Combining the real and imaginary parts gives <math>\frac{n}{2} + 1 - \frac{n}{2}i</math>, which is equivalent to option <math>\textbf{(C)}</math>. | ||
==See Also== | ==See Also== |
Revision as of 02:49, 25 July 2019
Problem
If is a multiple of , the sum , where , equals:
Solution
The real part is , which is . If is a multiple of , then we have an odd number of terms in total: we start with at , then add two more terms to get at , etc. With each successive addition, we're really adding a total of , since , and , etc.
At , the sum is , and at , the sum is . Since the sum increases linearly, the real part of the sum is .
The imaginary part is , which is . This time, we have an even number of terms. We group pairs of terms to get , and notice that each pair gives . Again, with the imaginary part is , while with the imaginary part is . Since again the sum increases linearly, this means the imaginary part is .
Combining the real and imaginary parts gives , which is equivalent to option .
See Also
1964 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 33 |
Followed by Problem 35 | |
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