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 $m+n$ .

Which is

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

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

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