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Difference between revisions of "2002 AIME I Problems/Problem 4"

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Revision as of 18:54, 4 July 2013

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,

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

Which is

$\dfrac{n-m}{mn}=\dfrac{1}{29}$

Since we need a 29 in the denominator, let $n=29t$.* Substituting,

$29t-m=mt$

$\frac{29t}{t+1} = m$

Since m is an integer, $t+1 = 29$, or $t=28$. It quickly follows that $n=29(28)$ and $m=28$, so $m+n = 30(28) = \fbox{840}$.

  • If $m=29t$, a similar argument to the one above implies $m=29(28)$ and $m=28$, which implies $m>n$, which is impossible since $n-m>0$.

See also

2002 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 3 		Followed by
Problem 5
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All AIME Problems and Solutions

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