2002 AMC 10P Problems/Problem 23

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.

ab=(121+223+325++100122001)(123+225+327++100122003)=(2212)3+(3222)5+42327++10012100022001+1100122003.

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,

ab=(2212)3+(3222)5+42327++10012100022001+1100122003=1001100122003>1001100122002=100110012=10012=500.5

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 (ProblemsAnswer KeyResources)
Preceded by
Problem 22
Followed by
Problem 24
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All AMC 10 Problems and Solutions

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