Difference between revisions of "1964 AHSME Problems/Problem 34"
Talkinaway (talk | contribs) |
Talkinaway (talk | contribs) |
||
Line 6: | Line 6: | ||
<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> | ||
+ | |||
+ | ==Solution== | ||
+ | |||
+ | 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. | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | 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>. | ||
==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 | |
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.