Question

Consider the sequence defined by $a_k = \frac{1}{k^2 + k}$ for $k \ge 1.$ Given that $a_m + a_{m+1} + \cdots + a_{n-1} = \frac{1}{29},$ for positive integers $m$ and $n$ with $m \lt n,$ find $m + n.$

Answer

Answer: 840

Concepts:telescoping partial fractions Diophantine Equation

Difficulty rating: 2110

Solution:

Since $\frac{1}{k^2 + k} = \frac{1}{k(k + 1)} = \frac{1}{k} - \frac{1}{k + 1},$ the sum telescopes: $\begin{aligned} &a_m + a_{m+1} + \cdots + a_{n-1} \\ &= \frac{1}{m} - \frac{1}{n} \\ &= \frac{1}{29}. \end{aligned}$

Multiplying through by $29mn$ gives $29n - 29m = mn,$ which rearranges to $(29 - m)(29 + n) = 29^2.$ Since $29$ is prime and $29 + n \gt 29,$ the only factorization with $m$ a positive integer is $29 - m = 1$ and $29 + n = 841,$ so $m = 28$ and $n = 812.$

Therefore $m + n = 28 + 812 = 840.$

2002 AIME I Problem 4

Answer: 840

Difficulty rating: 2110

Solution:

Problem 4 in Other Years