Difference between revisions of "2002 AMC 12P Problems/Problem 14"
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Note that <math>i^4 = 1</math>, so <math>i^n = i^{4m+n}</math> for all integers <math>m</math> and <math>n</math>. In particular, <math>i = 1</math>, <math>i^2 = -1</math>, and <math>i^3 = -i</math>. | Note that <math>i^4 = 1</math>, so <math>i^n = i^{4m+n}</math> for all integers <math>m</math> and <math>n</math>. In particular, <math>i = 1</math>, <math>i^2 = -1</math>, and <math>i^3 = -i</math>. | ||
We group the positive and negative real terms together and group the positive and negative imaginary parts together. | We group the positive and negative real terms together and group the positive and negative imaginary parts together. | ||
Latest revision as of 20:07, 10 March 2024
Problem
Find 
Solution 1
Note that 
, so 
for all integers 
and 
. In particular, 
, 
, and 
.
We group the positive and negative real terms together and group the positive and negative imaginary parts together.
The positive real terms have exponents on 
that are multiples of 4. Therefore, the positive real part evaluates to
![\[4 + 8 + ... + 2000\]](http://latex.artofproblemsolving.com/e/f/c/efcbdd533f8637bc8aedec697893d8204968e3f9.png)
The negative real terms have exponents on that are of the form 
for integers 
. Therefore, the negative real part evaluates to ![\[-(2 + 6 + ... + 2002)\]](http://latex.artofproblemsolving.com/0/0/b/00bd687011cf3c5c5a5020f598a38fa174bb4e1b.png)
The positive imaginary terms have exponents on that are of the form 
for integers . Therefore, the negative real part evaluates to ![\[(1 + 5 + ... + 2001)i\]](http://latex.artofproblemsolving.com/7/5/4/754999b87401e887f12f85ac44a740e8e917f1d8.png)
The negative imaginary terms have exponents on that are of the form 
for integers . Therefore, the negative real part evaluates to ![\[-(3 + 7 + ... + 1999)i\]](http://latex.artofproblemsolving.com/9/7/2/972b7c22aca8bb4db4e90ea7f5f0fc17ff2bffc7.png)
Putting everything together, we have ![\[i + 2i^2 + ... + 2002i^{2002} = (-2 + 4 - 6 + 8 - ... + 2000 - 2002) + (1 - 3 + 5 - 7 + ... - 1999 + 2001)i\]](http://latex.artofproblemsolving.com/e/f/0/ef04ed085bbd0e3f7f15e68d42287399b38e7edb.png)
Group every 2 consecutive terms as shown below ![\[((-2 + 4)+(-6 + 8) + ... + (-1998 + 2000)-2002) + ((1 - 3)+(5 - 7) + ... + (1997 - 1999) + 2001)i\]](http://latex.artofproblemsolving.com/5/f/a/5fada0491bbe457cd63ee63f8b346b10c3192e3a.png)
Now we evaluate each small bracket with 2 terms. We get 
in the real part and 
in the imaginary part. Therefore, the sum becomes 
.
Note: This problem is similar to https://artofproblemsolving.com/wiki/index.php/2009_AMC_12A_Problems/Problem_15
See also
| 2002 AMC 12P (Problems • Answer Key • Resources) | |
| Preceded by Problem 13 |
Followed by Problem 15 |
| 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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
