《2026年点金训练精讲巧练高中数学选择性必修第二册人教A版》

思考 2. 将数列$\{ a_{n}\}$满足的条件改为“$a_{1}=1$，$a_{2}=3$，且$2na_{n}=(n - 1)a_{n - 1}+(n + 1)a_{n + 1}$（$n\geqslant 2$）”，能判断哪个数列是等差数列？如何进一步求数列$\{ a_{n}\}$的通项公式？

答案： 思考2.提示:能判断数列$\{na_n\}$是等差数列.
因为$2na_n = (n - 1)a_{n - 1} + (n + 1)a_{n + 1}(n \geq 2)$,
所以$na_n - (n - 1)a_{n - 1} = (n + 1)a_{n + 1} - na_n(n \geq 2)$.
故数列$\{(n + 1)a_{n + 1} - na_n\}$为常数列,且$2a_2 - a_1 = 5$,
所以$\{(n + 1)a_{n + 1} - na_n\} = 5$,
因此数列$\{na_n\}$是以$1$为首项,$5$为公差的等差数列.
所以$na_n = 1 + 5(n - 1) = 5n - 4$,因此$a_n = \frac{5n - 4}{n}$.

1. 已知数列$\{ a_{n}\}$满足$a_{1}=2$，$a_{n + 1}=\dfrac {2a_{n}}{a_{n}+2}$.
(1) 数列$\left\{ \dfrac {1}{a_{n}}\right\}$是否为等差数列？说明理由.
(2) 求$a_{n}$.

答案： 1.解:
(1)数列$\{\frac{1}{a_n}\}$是等差数列.理由如下:
因为$a_1 = 2$,$a_{n + 1} = \frac{2a_n}{a_n + 2}$,显然$a_n ＞ 0$,
所以$\frac{1}{a_{n + 1}} = \frac{a_n + 2}{2a_n} = \frac{1}{2} + \frac{1}{a_n}$,即$\frac{1}{a_{n + 1}} - \frac{1}{a_n} = \frac{1}{2}$.
所以数列$\{\frac{1}{a_n}\}$是首项为$\frac{1}{2}$,公差为$\frac{1}{2}$的等差数列.
(2)由
(1)可知$\frac{1}{a_n} = \frac{1}{2} + \frac{1}{2}(n - 1) = \frac{n}{2}$,所以$a_n = \frac{2}{n}$.

2. 已知数列$\{ a_{n}\}$，$a_{1}=\dfrac {3}{5}$，$a_{n}=2-\dfrac {1}{a_{n - 1}}$（$n\geqslant 2$，$n\in \mathbf{N}^{*}$），数列$\{ b_{n}\}$满足$b_{n}=\dfrac {1}{a_{n}-1}$（$n\in \mathbf{N}^{*}$）. 求证：数列$\{ b_{n}\}$是等差数列，并求出其通项公式.

答案： 2.证明:因为$a_n = 2 - \frac{1}{a_{n - 1}}(n \geq 2,n \in \mathbf{N}^*)$,$b_n = \frac{1}{a_n - 1}(n \in \mathbf{N}^*)$,
所以$b_{n + 1} - b_n = \frac{1}{a_{n + 1} - 1} - \frac{1}{a_n - 1}$
$= \frac{1}{2 - \frac{1}{a_n} - 1} - \frac{1}{a_n - 1}$
$= \frac{a_n}{a_n - 1} - \frac{1}{a_n - 1} = 1$.
又$b_1 = \frac{1}{a_1 - 1} = \frac{5}{2}$,
所以数列$\{b_n\}$是以$\frac{5}{2}$为首项,$1$为公差的等差数列,
通项公式为$b_n = n - \frac{7}{2}$.