2002 AIME I Problems/Problem 4

Revision as of 16:13, 18 August 2008 by Duelist (talk | contribs) (Solution)

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

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

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

$29t-n=nt$ $\frac{29t}{t+1} = n$

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 $m-n>0$.

See also

2002 AIME I (ProblemsAnswer KeyResources)
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