Difference between revisions of "2002 AMC 12P Problems/Problem 14"
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The negative imaginary terms have exponents on <math>i</math> that are of the form <math>4k + 3</math> for integers <math>k</math>. Therefore, the negative real part evaluates to <cmath>-(3 + 7 + ... + 1999)i</cmath> | The negative imaginary terms have exponents on <math>i</math> that are of the form <math>4k + 3</math> for integers <math>k</math>. Therefore, the negative real part evaluates to <cmath>-(3 + 7 + ... + 1999)i</cmath> | ||
| − | Putting everything together, we have <math>i + 2i^2 + ... + 2002i^{2002} = (-2+4-6+8- ... +2000-2002) + (1-3+5-7+ ... -1999+2001)i</math>. | + | Putting everything together, we have <math>i + 2i^2 + ... + 2002i^{2002} = (-2 + 4 - 6 + 8 - ... + 2000 - 2002) + (1 - 3 + 5 - 7 + ... - 1999 + 2001)i</math>. |
| − | Group every 2 consecutive terms as shown below <cmath>((-2+4)+(-6+8)+ ... +(-1998+2000)-2002) + ((1-3)+(5-7)+ ... +(1997-1999)+2001)i</cmath> | + | Group every 2 consecutive terms as shown below <cmath>((-2 + 4)+(-6 + 8) + ... + (-1998 + 2000)-2002) + ((1 - 3)+(5 - 7) + ... + (1997 - 1999) + 2001)i</cmath> |
Now we evaluate each small bracket with 2 terms. We get <math>500(2) = 1000</math> in the real part and <math>500(-2) = -1000</math> in the imaginary part. Therefore, the sum becomes <math>(1000 - 2002) + (-1000 + 2001)i = \boxed {\text{(D) }-1002 + 1001i}</math>. | Now we evaluate each small bracket with 2 terms. We get <math>500(2) = 1000</math> in the real part and <math>500(-2) = -1000</math> in the imaginary part. Therefore, the sum becomes <math>(1000 - 2002) + (-1000 + 2001)i = \boxed {\text{(D) }-1002 + 1001i}</math>. | ||
Revision as of 19:42, 10 March 2024
Problem
Find 
Solution
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 
.
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 
.
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. 
