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

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== Solution ==
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== 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.
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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>.
  
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Note: This problem is similar to https://artofproblemsolving.com/wiki/index.php/2009_AMC_12A_Problems/Problem_15
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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

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