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)
(→‎Solution 1)
 
(13 intermediate revisions by 2 users not shown)
Line 14: Line 14:
 
</math>
 
</math>
  
−
== Solution ==
+
== Solution 1 ==
 
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.
 +
 
The positive real terms have exponents on <math>i</math> that are multiples of 4. Therefore, the positive real part evaluates to  
 
The positive real terms have exponents on <math>i</math> that are multiples of 4. Therefore, the positive real part evaluates to  
 
<cmath>4 + 8 + ... + 2000</cmath>
 
<cmath>4 + 8 + ... + 2000</cmath>
Line 23: Line 24:
 
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
+
Putting everything together, we have <cmath>i + 2i^2 + ... + 2002i^{2002} = (-2 + 4 - 6 + 8 - ... + 2000 - 2002) + (1 - 3 + 5 - 7 + ... - 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>.
 +
 
 +
Note: This problem is similar to [[2009 AMC 12A Problems/Problem 15|2009 AMC 12A Problem 15]]
  
 
== 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}}

Latest revision as of 08:10, 15 July 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 1

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 \[i + 2i^2 + ... + 2002i^{2002} = (-2 + 4 - 6 + 8 - ... + 2000 - 2002) + (1 - 3 + 5 - 7 + ... - 1999 + 2001)i\]

Group every 2 consecutive terms as shown below \[((-2 + 4)+(-6 + 8) + ... + (-1998 + 2000)-2002) + ((1 - 3)+(5 - 7) + ... + (1997 - 1999) + 2001)i\]

Now we evaluate each small bracket with 2 terms. We get $500(2) = 1000$ in the real part and $500(-2) = -1000$ in the imaginary part. Therefore, the sum becomes $(1000 - 2002) + (-1000 + 2001)i = \boxed {\text{(D) }-1002 + 1001i}$.

Note: This problem is similar to 2009 AMC 12A 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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2002_AMC_12P_Problems/Problem_14&oldid=222570"