《2025年优化探究同步导学案高中数学选择性必修第二册人教版》

[例 1] 已知数列$\{ a_{n}\}$满足$a_{1} = 2,a_{n + 1} = \frac{2a_{n}}{a_{n} + 2}$.
(1)数列$\{\frac{1}{a_{n}}\}$是否为等差数列？若是，请说明理由.
(2)求$a_{n}$.
[分析] 要判断数列$\{\frac{1}{a_{n}}\}$是否为等差数列，关键看$\frac{1}{a_{n + 1}} - \frac{1}{a_{n}}$是否为常数.
课堂笔记：
[变条件] 将本例中的条件$``a_{1} = 2,a_{n + 1} = \frac{2a_{n}}{a_{n} + 2}$''，换为$``a_{1} = 4,a_{n} = 4 - \frac{4}{a_{n - 1}}(n ＞ 1)$，记$b_{n} = \frac{1}{a_{n} - 2}$''.
(1)试证明数列$\{ b_{n}\}$为等差数列；
(2)求数列$\{ a_{n}\}$的通项公式.

答案： [解]
(1)数列$\left \{ \frac{1}{a_{n}} \right \}$是等差数列,理由如下:因为$a_{1}=2,a_{n + 1}=\frac{2a_{n}}{a_{n} + 2}$,所以$\frac{1}{a_{n + 1}} = \frac{a_{n} + 2}{2a_{n}} = \frac{1}{2} + \frac{1}{a_{n}}$,所以$\frac{1}{a_{n + 1}} - \frac{1}{a_{n}} = \frac{1}{2}$,即$\frac{1}{a_{n}}$是首项为$\frac{1}{a_{1}} = \frac{1}{2}$,公差为$d = \frac{1}{2}$的等差数列.
(2)由
(1)可知$\frac{1}{a_{n}} = \frac{1}{a_{1}} + (n - 1)d = \frac{n}{2}$,所以$a_{n} = \frac{2}{n}$.
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(1)证明:$b_{n + 1} - b_{n} = \frac{1}{a_{n + 1} - 2} - \frac{1}{a_{n} - 2}$$=( \frac{4}{a_{n}} - 2) - \frac{1}{a_{n} - 2} = \frac{a_{n} - 2}{2(a_{n} - 2)} = \frac{1}{2}$.又$b_{1} = \frac{1}{a_{1} - 2} = - \frac{1}{2}$,$\therefore$数列$\{ b_{n}\}$是首项为$\frac{1}{2}$,公差为$\frac{1}{2}$的等差数列.
(2)解:由
(1)知$b_{n} = \frac{1}{2} + (n - 1) × \frac{1}{2}$$=\frac{n}{2}$.$\because b_{n} = \frac{a_{n} - 2}{2} ·s a_{n} = 2b_{n} + 2 = \frac{2}{n} + 2$,$\therefore$数列$\{ a_{n}\}$的通项公式为$a_{n} = \frac{2}{n} + 2$,$n \in N^{*}$.

1. 已知数列$\{ a_{n}\}$的前$n$项和为$S_{n} = n^{2} + 2n$.
(1)求数列$\{ a_{n}\}$的通项公式；
(2)求证：数列$\{ a_{n}\}$是等差数列.

答案：
(1)解:由$n \in N^{*},S_{n} = n^{2} + 2n$,
得当$n \geq 2$时,$S_{n - 1} = (n - 1)^{2} + 2(n - 1)$,
于是$a_{n} = S_{n} - S_{n - 1} = n^{2} + 2n - [(n - 1)^{2} + 2(n - 1)] = 2n + 1$,
而当$n = 1$时,$a_{1} = S_{1} = 1^{2} + 2 × 1 = 3$亦满足上式,
所以数列$\{ a_{n}\}$的通项公式为$a_{n} = 2n + 1,n \in N^{*}$.
(2)证明:由
(1)知,$a_{n} = 2n + 1$,
当$n \geq 2$时,
$a_{n - 1} = 2(n - 1) + 1 = 2n - 1$,
因此$a_{n} - a_{n - 1} = 2n + 1 - (2n - 1) = 2$.
所以数列$\{ a_{n}\}$是一个以2为公差的等差数列.