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

m (→‎Solution)
m (→‎Solution)
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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 + ... + 2020</cmath>
+
<cmath>4 + 8 + ... + 2000</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 + ... + 2022)</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 + ... + 2021)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>
  
 
== 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 19:27, 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 \[(1 + 5 + ... + 2001)i\]

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

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