Question

Considere la sucesión definida por $a_k = \frac{1}{k^2 + k}$ para $k \ge 1.$ Dado que $a_m + a_{m+1} + \cdots + a_{n-1} = \frac{1}{29},$ para enteros positivos $m$ y $n$ con $m \lt n,$ halle $m + n.$

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 $m + n.$

Respuesta

Solución:

Como $\frac{1}{k^2 + k} = \frac{1}{k(k + 1)} = \frac{1}{k} - \frac{1}{k + 1},$ la suma telescopa: $\begin{aligned} &a_m + a_{m+1} + \cdots + a_{n-1} \\ &= \frac{1}{m} - \frac{1}{n} \\ &= \frac{1}{29}. \end{aligned}$

Multiplicando todo por $29mn$ se obtiene $29n - 29m = mn,$ que se reordena como $(29 - m)(29 + n) = 29^2.$ Como $29$ es primo y $29 + n \gt 29,$ la única factorización con $m$ entero positivo es $29 - m = 1$ y $29 + n = 841,$ así que $m = 28$ y $n = 812.$

Por lo tanto $m + n = 28 + 812 = 840.$

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 Problema 4

Solución:

El Problema 4 en otros años