Difference between revisions of "1964 AHSME Problems/Problem 34"
Talkinaway (talk | contribs) |
Talkinaway (talk | contribs) (→Solution) |
||
Line 15: | Line 15: | ||
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>. | 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>. | ||
− | 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>. | + | Combining the real and imaginary parts gives <math>\frac{n}{2} + 1 - \frac{n}{2}i</math>, which is equivalent to option <math>\boxed{\textbf{(C)}}</math>. |
==See Also== | ==See Also== |
Latest revision as of 01:49, 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 | |
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 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 | ||
All AHSME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.