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

(Solution)
(Solution)
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== Solution ==
 
== Solution ==
This problem is literally almost an exact copy of [[2009 AMC 12A Problems/Problem 15|2009 AMC 12A Problem 15]], or arguably easier.
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This problem is literally almost an exact copy of [[2009 AMC 12A Problems/Problem 15|2009 AMC 12A Problem 15]], or arguably easier. Since I'm too lazy to copy over the other solutions which all work, I'll just copy the general idea of Solutions <math>1</math> and <math>2</math> from there, since that's how I solved the problem on both tests. Cycling
  
 
== 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 01:55, 31 December 2023

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

This problem is literally almost an exact copy of 2009 AMC 12A Problem 15, or arguably easier. Since I'm too lazy to copy over the other solutions which all work, I'll just copy the general idea of Solutions $1$ and $2$ from there, since that's how I solved the problem on both tests. Cycling

See also

2002 AMC 12P (ProblemsAnswer KeyResources)
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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