25~26高三上·云南11月联考·第14题

$1220$

【题目】数列${a_n}$满足：$a_1 = \frac{2}{3}$，且$a_{n+1} – a_n = \sqrt{\frac{2}{3}(a_{n+1} + a_n)}$，则$a_{60} =$_______．

【答案】$1220$

【解析】由$a_{n+1} – a_n = \sqrt{\frac{2}{3}(a_{n+1} + a_n)}$，
所以$(a_{n+1} – a_n)^2 = \frac{2}{3}(a_{n+1} + a_n)$，
且$(a_{n+2} – a_{n+1})^2 = \frac{2}{3}(a_{n+2} + a_{n+1})$，
两式相减得：$(a_{n+2} – a_n)(a_{n+2} – 2a_{n+1} + a_n) = \frac{2}{3}(a_{n+2} – a_n)$，
又由$a_{n+1} – a_n = \sqrt{\frac{2}{3}(a_{n+1} + a_n)} \geq 0$及$a_1 > 0$，
故${a_n}$是递增数列，$a_{n+2} – a_n \neq 0$，
所以$a_{n+2} – a_{n+1} = a_{n+1} – a_n + \frac{2}{3}$，

当$n = 1$时，$a_2 – a_1 = \sqrt{\frac{2}{3}(a_2 + a_1)}$，解得$a_2^2 = 2a_2$，
又$a_2 > a_1$，即$a_2 = 2$，$a_2 – a_1 = \frac{4}{3}$，

所以数列${a_{n+1} – a_n}$是等差数列，首项为$\frac{4}{3}$，公差为$\frac{2}{3}$，

所以$a_{n+1} – a_n = \frac{4}{3} + (n – 1) \times \frac{2}{3} = \frac{2}{3}(n + 1)$，

故$a_{60} = (a_{60} – a_{59}) + (a_{59} – a_{58}) + \cdots + (a_2 – a_1) + a_1 $$= \frac{2}{3}(2 + 3 + \cdots + 60) + \frac{2}{3}$
$= \frac{2}{3} \times \frac{59 \times (2 + 60)}{2} + \frac{2}{3} = 1220$.

故答案为：$1220$．