《2025年高考总复习优化设计高中数学北师大版》

例2 (2023·新高考Ⅱ，18)已知$\{ a_{n}\}$为等差数列，$b_{n}=\left\{\begin{array}{l} a_{n}-6,n 为奇数,\\ 2a_{n},n 为偶数.\end{array}\right.$记$S_{n},T_{n}$分别为数列$\{ a_{n}\},\{ b_{n}\}$的前$n$项和，$S_{4}=32,T_{3}=16.$
(1)求$\{ a_{n}\}$的通项公式；
(2)证明：当$n＞5$时，$T_{n}＞S_{n}.$

答案：
(1)解 设等差数列\{a_n\}的公差为d。
由$b_n = \begin{cases}a_n - 6,n$为奇数,\\2a_n,n为偶数$,\end{cases}$得$b_1 = a_1 - 6，$
$b_2 = 2a_2 = 2(a_1 + d)，$$b_3 = a_3 - 6 = a_1 + 2d - 6。$
则由$S_4 = 32,T_3 = 16，$得$\begin{cases}4a_1 + \frac{4×3}{2}xd = 32,\a_1 - 6) + 2(a_1 + d) + (a_1 + 2d - 6) = 16,\end{cases}$
解得$\begin{cases}a_1 = 5,\\d = 2.\end{cases}$
所以$a_n = a_1 + (n - 1)d = 2n + 3。$
(2)证明 由
(1)可得$S_n = \frac{n[5 + (2n + 3)]}{2} = \frac{n^2 + 4n}{2}。$
当n为奇数时，$T_n = a_1 - 6 + 2a_2 + a_3 - 6 + 2a_4 + a_5 - 6 + 2a_6 + \cdots + a_{n - 2} - 6 + 2a_{n - 1} + a_n - 6 = (-1 + 14) + (3 + 22) + (7 + 30) + \cdots + [(2n - 7) + (4n + 2)] + 2n - 3 = [-1 + 3 + \cdots + (2n - 7)] + [14 + 22 + \cdots + (4n + 2)] = \frac{\frac{n - 1}{2}(-1 + 2n - 3)}{2} + \frac{\frac{n - 1}{2}(14 + 4n + 2)}{2} = \frac{3n^2 + 5n - 10}{2}。$
当n ＞ 5时，$T_n - S_n = \frac{3n^2 + 5n - 10}{2} - \frac{n^2 + 4n}{2} = \frac{n^2 - 3n - 10}{(n - 5)(n + 2)} ＞ 0，$所以T_n ＞ S_n。
当n为偶数时，$T_n = a_1 - 6 + 2a_2 + a_3 - 6 + 2a_4 + a_5 - 6 + 2a_6 + \cdots + a_{n - 1} - 6 + 2a_n = (-1 + 14) + (3 + 22) + (7 + 30) + \cdots + [(2n - 5) + (4n + 6)] = [-1 + 3 + \cdots + (2n - 5)] + [14 + 22 + \cdots + (4n + 6)] = \frac{\frac{n}{2}(-1 + 2n - 5)}{2} + \frac{\frac{n}{2}(14 + 4n + 6)}{2} = \frac{3n^2 + 7n}{2}。$
当n ＞ 5时，$T_n - S_n = \frac{3n^2 + 7n}{2} - (n^2 + 4n) = \frac{n^2 - n}{2} = \frac{n(n - 1)}{2} ＞ 0，$所以T_n ＞ S_n。
综上可知，当n ＞ 5时，T_n ＞ S_n。

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答案： 答案略

[对点训练2](2024·山东师大附中模拟)已知$\{ a_{n}\}$是各项均为正数的数列，$S_{n}$为$\{ \sqrt {a_{n}}\}$的前$n$项和，且$\sqrt {a_{n}},S_{n},a_{n}-2$成等差数列.
(1)求$\{ a_{n}\}$的通项公式；
(2)已知$b_{n}=(-1)^{n}a_{n}$，求数列$\{ b_{n}\}$的前$n$项和$T_{n}.$
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答案：
(1)由$\sqrt{a_n},S_n,a_n - 2$成等差数列，得$2S_n = \sqrt{a_n} + a_n - 2$，①
当$n = 1$时，$2\sqrt{a_1} = \sqrt{a_1} + a_1 - 2$，
$\therefore a_1 - \sqrt{a_1} - 2 = 0$，得$\sqrt{a_1} = 2(\sqrt{a_1} = -1$舍去)，
当$n\geq2$时，$2S_{n - 1} = \sqrt{a_{n - 1}} + a_{n - 1} - 2$，②
① - ②得$2\sqrt{a_n} = \sqrt{a_n} - \sqrt{a_{n - 1}} + a_n - a_{n - 1}$，$\therefore\sqrt{a_n} + \sqrt{a_{n - 1}} = a_n - a_{n - 1} = (\sqrt{a_n} + \sqrt{a_{n - 1}})(\sqrt{a_n} - \sqrt{a_{n - 1}})$。
又$\sqrt{a_n} + \sqrt{a_{n - 1}} \neq 0$，
$\therefore\sqrt{a_n} - \sqrt{a_{n - 1}} = 1$，
$\therefore\{\sqrt{a_n}\}$是首项为$2$，公差为$1$的等差数列，
$\therefore\sqrt{a_n} = 2 + n - 1 = n + 1$，
故$a_n = (n + 1)^2$。
(2)由
(1)知$b_n = (-1)^n(n + 1)^2$，
当$n$是奇数时，$T_n = -2^2 + 3^2 - 4^2 + 5^2 - 6^2 + 7^2 - \cdots - (n - 1)^2 + n^2 - (n + 1)^2 = (3 - 2)×(3 + 2) + (5 - 4)×(5 + 4) + (7 - 6)×(7 + 6) + \cdots + [n - (n - 1)][n + (n - 1)] - (n + 1)^2 = \frac{5 + 2n - 1}{2}×\frac{n - 1}{2} - (n + 1)^2 = \frac{-n^2 - 3n - 4}{2}$；
当$n$是偶数时，$T_n = -2^2 + 3^2 - 4^2 + 5^2 - 6^2 + 7^2 - \cdots - n^2 + (n + 1)^2 = (3 - 2)×(3 + 2) + (5 - 4)×(5 + 4) + (7 - 6)×(7 + 6) + \cdots + [(n + 1) - n][(n + 1) + n] = 5 + 9 + 13 + \cdots + (2n + 1) = \frac{5 + 2n + 1}{2}×\frac{n}{2} = \frac{n^2 + 3n}{2}$。
综上，$T_n = \begin{cases}\frac{-n^2 - 3n - 4}{2},n为奇数,\frac{n^2 + 3n}{2},n为偶数.\end{cases}$