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 "2002 AMC 10P Problems/Problem 23"

(→‎Solution 1)
m (→‎Solution 1)
 
(6 intermediate revisions by one other user not shown)
Line 26: Line 26:
  
 
\begin{align*}
 
\begin{align*}
−
<math>a-b
+
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{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}.</math>
+
&=\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*}
 
\end{align*}
 +
 +
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,
 +
 +
\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 <math>a-b>500.5,</math> we can conclude that <math>a-b</math> is closer to <math>501</math> than <math>500.</math>
 +
 +
Thus, our answer is  <math>\boxed{\textbf{(B) } 501}.</math>
  
 
== See also ==
 
== See also ==
 
{{AMC10 box|year=2002|ab=P|num-b=22|num-a=24}}
 
{{AMC10 box|year=2002|ab=P|num-b=22|num-a=24}}
 
{{MAA Notice}}
 
{{MAA Notice}}

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

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=2002_AMC_10P_Problems/Problem_23&oldid=222810"