Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2002 AMC 12P Problems/Problem 14"

m (→‎Solution)
m (→‎Solution)
Line 21: Line 21:
 
The negative real terms have exponents on <math>i</math> that are of the form <math>4k + 2</math> for integers <math>k</math>. Therefore, the negative real part evaluates to <cmath>-(2 + 6 + ... + 2002)</cmath>
 
The negative real terms have exponents on <math>i</math> that are of the form <math>4k + 2</math> for integers <math>k</math>. Therefore, the negative real part evaluates to <cmath>-(2 + 6 + ... + 2002)</cmath>
 
The positive imaginary terms have exponents on <math>i</math> that are of the form <math>4k + 1</math> for integers <math>k</math>. Therefore, the negative real part evaluates to <cmath>(1 + 5 + ... + 2001)i</cmath>
 
The positive imaginary terms have exponents on <math>i</math> that are of the form <math>4k + 1</math> for integers <math>k</math>. Therefore, the negative real part evaluates to <cmath>(1 + 5 + ... + 2001)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
  
 
== See also ==
 
== See also ==
 
{{AMC12 box|year=2002|ab=P|num-b=13|num-a=15}}
 
{{AMC12 box|year=2002|ab=P|num-b=13|num-a=15}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 18:28, 10 March 2024

Problem

Find $i + 2i^2 +3i^3 + ... + 2002i^{2002}.$

$\text{(A) }-999 + 1002i \qquad \text{(B) }-1002 + 999i \qquad \text{(C) }-1001 + 1000i \qquad \text{(D) }-1002 + 1001i \qquad \text{(E) }i$

Solution

Note that $i^4 = 1$, so $i^n = i^{4m+n}$ for all integers $m$ and $n$. In particular, $i = 1$, $i^2 = -1$, and $i^3 = -i$. 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 $i$ that are multiples of 4. Therefore, the positive real part evaluates to \[4 + 8 + ... + 2000\] The negative real terms have exponents on $i$ that are of the form $4k + 2$ for integers $k$. Therefore, the negative real part evaluates to \[-(2 + 6 + ... + 2002)\] The positive imaginary terms have exponents on $i$ that are of the form $4k + 1$ for integers $k$. Therefore, the negative real part evaluates to \[(1 + 5 + ... + 2001)i\] The negative imaginary terms have exponents on $i$ that are of the form $4k + 3$ for integers $k$. Therefore, the negative real part evaluates to \[-(3 + 7 + ... + 1999)i\]

Putting everything together, we have

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2002_AMC_12P_Problems/Problem_14&oldid=216813"