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Difference between revisions of "2002 AMC 10P Problems/Problem 23"

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Notice how <math>\frac{(2^2-1^2)}{3}=1, \frac{(3^2-2^2)}{5}=1, \frac{4^2-3^2}{7}=1,</math> etc. This is because all of these are in the form <math>\frac{n^2-(n-1)^2}{2n-1}=\frac{n^2-(n^2-2n+1)}{2n-1}=\frac{2n-1}{2n-1}=1</math>. There are <math>1001</math> of these terms since it begins at <math>n=2</math> and ends at <math>n=1001,</math> so <math>1001-2+1=1000.</math> Therefore, <math>a-b=1000+1 - \frac{1001^2}{2003}.</math> We can either manually calculate <math>\frac{1001^2}{2003}</math> or notice that <math>\frac{1001^2}{2003} \approx \frac{1001^2}{2002}.</math> <math>\frac{1001^2}{2003} < \frac{1001^2}{2002}</math>, so <math>-\frac{1001^2}{2003} > -\frac{1001^2}{2002}.</math> Therefore,
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Notice how <math>\frac{(2^2-1^2)}{3}=1, \frac{(3^2-2^2)}{5}=1, \frac{4^2-3^2}{7}=1, \; \dots \;, \frac{1001^2-1000^2}{2001}=1</math>. This is because all of these are in the form <math>\frac{n^2-(n-1)^2}{2n-1}=\frac{n^2-(n^2-2n+1)}{2n-1}=\frac{2n-1}{2n-1}=1</math>. There are <math>1000</math> of these terms since it begins at <math>n=2</math> and ends at <math>n=1001,</math> so <math>1001-2+1=1000.</math> Therefore, <math>a-b=1000+1 - \frac{1001^2}{2003}.</math> We can either manually calculate <math>\frac{1001^2}{2003}</math> or notice that <math>\frac{1001^2}{2003} \approx \frac{1001^2}{2002}.</math> <math>\frac{1001^2}{2003} < \frac{1001^2}{2002}</math>, so <math>-\frac{1001^2}{2003} > -\frac{1001^2}{2002}.</math> Therefore,
  
 
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Latest revision as of 12:49, 16 July 2024

Problem

Let \[a=\frac{1^2}{1} + \frac{2^2}{3} + \frac{3^2}{5} + \; \dots \; + \frac{1001^2}{2001}\]

and

\[b=\frac{1^2}{3} + \frac{2^2}{5} + \frac{3^2}{7} + \; \dots \; + \frac{1001^2}{2003}.\]

Find the integer closest to $a-b.$

$\text{(A) }500 \qquad \text{(B) }501 \qquad \text{(C) }999 \qquad \text{(D) }1000 \qquad \text{(E) }1001$

Solution 1

Start by subtracting $a$ and $b$ and group those with a common denominator together, leaving $\frac{1^2}{1}$ and $\frac{1001^2}{2003}$ to the side.

\begin{align*} a-b &=(\frac{1^2}{1} + \frac{2^2}{3} + \frac{3^2}{5} + \; \dots \; + \frac{1001^2}{2001}) - (\frac{1^2}{3} + \frac{2^2}{5} + \frac{3^2}{7} + \; \dots \; + \frac{1001^2}{2003}) \\ &=\frac{(2^2-1^2)}{3}+\frac{(3^2-2^2)}{5}+\frac{4^2-3^2}{7} + \; \dots \; + \frac{1001^2-1000^2}{2001} + 1 - \frac{1001^2}{2003}. \\ \end{align*}

Notice how $\frac{(2^2-1^2)}{3}=1, \frac{(3^2-2^2)}{5}=1, \frac{4^2-3^2}{7}=1, \; \dots \;, \frac{1001^2-1000^2}{2001}=1$. This is because all of these are in the form $\frac{n^2-(n-1)^2}{2n-1}=\frac{n^2-(n^2-2n+1)}{2n-1}=\frac{2n-1}{2n-1}=1$. There are $1000$ of these terms since it begins at $n=2$ and ends at $n=1001,$ so $1001-2+1=1000.$ Therefore, $a-b=1000+1 - \frac{1001^2}{2003}.$ We can either manually calculate $\frac{1001^2}{2003}$ or notice that $\frac{1001^2}{2003} \approx \frac{1001^2}{2002}.$ $\frac{1001^2}{2003} < \frac{1001^2}{2002}$, so $-\frac{1001^2}{2003} > -\frac{1001^2}{2002}.$ Therefore,

\begin{align*} a-b &=\frac{(2^2-1^2)}{3}+\frac{(3^2-2^2)}{5}+\frac{4^2-3^2}{7} + \; \dots \; + \frac{1001^2-1000^2}{2001} + 1 - \frac{1001^2}{2003} \\ &=1001-\frac{1001^2}{2003} \\ &>1001-\frac{1001^2}{2002} \\ &=1001-\frac{1001}{2} \\ &=\frac{1001}{2} \\ &=500.5 \\ \end{align*}

Since $a-b>500.5,$ we can conclude that $a-b$ is closer to $501$ than $500.$

Thus, our answer is $\boxed{\textbf{(B) } 501}.$

See also

2002 AMC 10P (Problems • Answer Key • Resources)
Preceded by
Problem 22 		Followed by
Problem 24
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 10 Problems and Solutions

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