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 1

$\dfrac{1}{k^2+k}=\dfrac{1}{k(k+1)}=\dfrac{1}{k}-\dfrac{1}{k+1}$ . 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 means that

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

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

$29t-m=mt$

so

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

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

*If , a similar argument to the one above implies  and , which implies . This is impossible since .

Solution 2

Note that $a_1 + a_2 + \cdots + a_i = \dfrac{i}{i+1}$ . This can be proven by induction. Thus, $\sum\limits_{i=m}^{n-1} a_i = \sum\limits_{i=1}^{n-1} a_i - \sum\limits_{i=1}^{m-1} a_i = \dfrac{n-1}{n} - \dfrac{m-1}{m} = \dfrac{n-m}{mn} = 1/29$ . Cross-multiplying yields $29n - 29m - mn = 0$ , and adding $29^2$ to both sides gives $(29-m)(29+n) = 29^2$ . Clearly, $m < n \implies 29 - m = 1$ and $29 + n = 29^2$ . Hence, $m = 28$ , $n = 812$ , and $m+n = \fbox{840}$ .

$~$ keeper1098

2002 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 3	Followed by Problem 5
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2002 AIME I Problems/Problem 4

Contents

Problem

Solution 1

Solution 2

See also