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Difference between revisions of "2009 AMC 12A Problems/Problem 15"

(New page: == Problem == For what value of <math>n</math> is <math>i + 2i^2 + 3i^3 + \cdots + ni^n = 48 + 49i</math>? Note: here <math>i = \sqrt { - 1}</math>. <math>\textbf{(A)}\ 24 \qquad \textbf...)
 
m (→‎Solution: Some LaTex typos)
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Obviously, even powers of <math>i</math> are real and odd powers of <math>i</math> are imaginary.  
 
Obviously, even powers of <math>i</math> are real and odd powers of <math>i</math> are imaginary.  
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Hence the real part of the sum is <math>2i^2 + 4i^4 + \6i^6 + cdots</math>, and  
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Hence the real part of the sum is <math>2i^2 + 4i^4 + 6i^6 + ldots</math>, and  
 
the imaginary part is <math>i + 3i^3 + 5i^5 + \cdots</math>.
 
the imaginary part is <math>i + 3i^3 + 5i^5 + \cdots</math>.
  
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We can rewrite the imaginary part as follows: <math>i + 3i^3 + 5i^5 + \cdots = i(1 + 3i^2 + 5i^4 + \cdots) = i(1 - 3 + 5 - \cdots)</math>. We need to obtain <math>(1 - 3 + 5 - \cdots) = 49</math>. 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 <math>1 + (-3+5) + (-7+9) + \cdots</math>. We need <math>24</math> parentheses, therefore the last value used is <math>97</math>. This happens when <math>n=97</math> or <math>n=98</math>, and we are done.
 
We can rewrite the imaginary part as follows: <math>i + 3i^3 + 5i^5 + \cdots = i(1 + 3i^2 + 5i^4 + \cdots) = i(1 - 3 + 5 - \cdots)</math>. We need to obtain <math>(1 - 3 + 5 - \cdots) = 49</math>. 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 <math>1 + (-3+5) + (-7+9) + \cdots</math>. We need <math>24</math> parentheses, therefore the last value used is <math>97</math>. This happens when <math>n=97</math> or <math>n=98</math>, and we are done.
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== See Also ==
 
== See Also ==
  
 
{{AMC12 box|year=2009|ab=A|num-b=14|num-a=16}}
 
{{AMC12 box|year=2009|ab=A|num-b=14|num-a=16}}

Revision as of 00:16, 20 February 2009

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

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
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