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 "2009 AMC 12A Problems/Problem 15"

(→‎Solution)
(→‎Solution 1)
Line 7: Line 7:
  
 
== Solution 1 ==  
 
== Solution 1 ==  
−
We know that <math>i^x</math> cycles every <math>4</math> numbers so we group the sum in <math>4</math>s.  
+
We know that <math>i^x</math> cycles every <math>4</math> powers so we group the sum in <math>4</math>s.  
 
<cmath>i+2i^2+3i^3+4i^4=2-2i</cmath>
 
<cmath>i+2i^2+3i^3+4i^4=2-2i</cmath>
 
<cmath>5i^5+6i^6+7i^7+8i^8=2-2i</cmath>
 
<cmath>5i^5+6i^6+7i^7+8i^8=2-2i</cmath>
Line 14: Line 14:
 
For 24 groups we thus, get <math>48-48i</math> as our sum.  
 
For 24 groups we thus, get <math>48-48i</math> as our sum.  
 
We know the solution must lie near
 
We know the solution must lie near
−
The next term is the <math>24*4+1=97</math>th term. This term is equal to <math>97i</math> (first in a group of <math>4</math>) and our sum is now <math>48+49i</math> so <math>n=97</math> is our answer
+
The next term is the <math>24*4+1=97</math>th term. This term is equal to <math>97i</math> (first in a group of <math>4</math> so <math>i^{97}=i</math>) and our sum is now <math>48+49i</math> so <math>n=97</math> is our answer
−
 
 
  
 
== Solution 2==
 
== Solution 2==

Revision as of 07:53, 8 March 2019

Problem

For what value of $n$ is $i + 2i^2 + 3i^3 + \cdots + ni^n = 48 + 49i$?

Note: here $i = \sqrt { - 1}$.

$\textbf{(A)}\ 24 \qquad \textbf{(B)}\ 48 \qquad \textbf{(C)}\ 49 \qquad \textbf{(D)}\ 97 \qquad \textbf{(E)}\ 98$

Solution 1

We know that $i^x$ cycles every $4$ powers so we group the sum in $4$s. \[i+2i^2+3i^3+4i^4=2-2i\] \[5i^5+6i^6+7i^7+8i^8=2-2i\]

We can postulate that every group of $4$ is equal to $2-2i$. For 24 groups we thus, get $48-48i$ as our sum. We know the solution must lie near The next term is the $24*4+1=97$th term. This term is equal to $97i$ (first in a group of $4$ so $i^{97}=i$) and our sum is now $48+49i$ so $n=97$ is our answer

Solution 2

Obviously, even powers of $i$ are real and odd powers of $i$ are imaginary. Hence the real part of the sum is $2i^2 + 4i^4 + 6i^6 + \ldots$, and the imaginary part is $i + 3i^3 + 5i^5 + \cdots$.

Let's take a look at the real part first. We have $i^2=-1$, hence the real part simplifies to $-2+4-6+8-10+\cdots$. If there were an odd number of terms, we could pair them as follows: $-2 + (4-6) + (8-10) + \cdots$, hence the result would be negative. As we need the real part to be $48$, we must have an even number of terms. If we have an even number of terms, we can pair them as $(-2+4) + (-6+8) + \cdots$. Each parenthesis is equal to $2$, thus there are $24$ of them, and the last value used is $96$. This happens for $n=96$ and $n=97$. As $n=96$ is not present as an option, we may conclude that the answer is $\boxed{97}$.

In a complete solution, we should now verify which of $n=96$ and $n=97$ will give us the correct imaginary part.

We can rewrite the imaginary part as follows: $i + 3i^3 + 5i^5 + \cdots = i(1 + 3i^2 + 5i^4 + \cdots) = i(1 - 3 + 5 - \cdots)$. We need to obtain $(1 - 3 + 5 - \cdots) = 49$. Once again we can repeat the same reasoning: If the number of terms were even, the left hand side would be negative, thus the number of terms is odd. The left hand side can then be rewritten as $1 + (-3+5) + (-7+9) + \cdots$. We need $24$ parentheses, therefore the last value used is $97$. This happens when $n=97$ or $n=98$, and we are done.

See Also

2009 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 14 		Followed by
Problem 16
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=2009_AMC_12A_Problems/Problem_15&oldid=104145"