2002 AIME I Problems/Problem 4

Problem

Consider the sequence defined by $a_k =\dfrac{1}{k^2+k}$ for $k\geq 1$ . Given that $a_m+a_{m+1}+\cdots+a_{n-1}=\dfrac{1}{29}$ , for positive integers $m$ and $n$ with $m<n$ , find $m+n$ .

Solution

$\dfrac{1}{k^2+k}=\dfrac{1}{k(k+1)}=\dfrac{1}{k}-\dfrac{1}{k+1}$ . Thus,

$a_m+a_{m+1}+a_{m+2}+\cdots +a_{n-1}=\dfrac{1}{m}-\dfrac{1}{m+1}+\dfrac{1}{m+1}-\dfrac{1}{m+2}+\cdots +\dfrac{1}{n-1}-\dfrac{1}{n-2}=\dfrac{1}{m}-\dfrac{1}{n-2}$

Which is

$\dfrac{n-2-m}{m(n-2)}=\dfrac{1}{29}$

We cross-multiply to get

$mn-2m=29n-58-29m\Rightarrow mn+27m-29n+58=0\Rightarrow (m-29)(n+27)=-725$

Thus $m$ is an integer less than 29. We try $m=4$ to get $n=2$ . $m=24, n=118$ . $m+n=\boxed{142}$

2002 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 3	Followed by Problem 5
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
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2002 AIME I Problems/Problem 4

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