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,

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

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 n 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 $n-m>0$ .

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
All AIME Problems and Solutions

Page

Toolbox

Search

2002 AIME I Problems/Problem 4

Problem

Solution

See also