《2025年创新设计高考总复习数学浙江专版》

例 1 已知数列$\{ a_{n}\}$的各项均为正数，记$S_{n}$为$\{ a_{n}\}$的前$n$项和，且$2S_{n}=a_{n}^{2}+a_{n}$。
(1)求数列$\{ a_{n}\}$的通项公式；
(2)记$c_{n}=(-1)^{n}a_{n}a_{n + 1}$，求数列$\{ c_{n}\}$的前$n$项和$T_{n}$。解 (1)当$n=1$时，$2S_1 = 2a_1 = a_1^2 + a_1$，解得$a_1 = 1$或$a_1 = 0$(舍去)，当$n \geq 2$时，由$2S_n = a_n^2 + a_n$得$2S_{n - 1} = a_{n - 1}^2 + a_{n - 1}$，两式相减得$2a_n = a_n^2 + a_n - a_{n - 1}^2 - a_{n - 1}$，$n \geq 2$，所以$(a_n + a_{n - 1})(a_n - a_{n - 1}) - (a_n + a_{n - 1}) = 0$，即$(a_n + a_{n - 1})(a_n - a_{n - 1} - 1) = 0$，又$\{a_n\}$各项均为正数，所以$a_n - a_{n - 1} = 1$，$n \geq 2$，所以$\{a_n\}$是公差为1的等差数列，所以$a_n = n$。(2)由(1)知$c_n = (-1)^n n(n + 1)$，当$n$为偶数时，$T_n = (-1) × 2 + 2 × 3 + (-3) × 4 + 4 × 5 + \cdots + [-(n - 1) × n] + n × (n + 1) = 2 × (-1 + 3) + 4 × (-3 + 5) + \cdots + n × [-(n - 1) + n + 1] = 2 × (2 + 4 + \cdots + n) = 2 × \frac{\frac{n}{2}(2 + n)}{2} = \frac{n(n + 2)}{2}$；

答案： 1. 当$n=1$时，$2S_1 = 2a_1 = a_1^2 + a_1$，解得$a_1 = 1$或$a_1 = 0$(舍去)，当$n \geq 2$时，由$2S_n = a_n^2 + a_n$得$2S_{n - 1} = a_{n - 1}^2 + a_{n - 1}$，两式相减得$2a_n = a_n^2 + a_n - a_{n - 1}^2 - a_{n - 1}$，$n \geq 2$，所以$(a_n + a_{n - 1})(a_n - a_{n - 1}) - (a_n + a_{n - 1}) = 0$，即$(a_n + a_{n - 1})(a_n - a_{n - 1} - 1) = 0$，又$\{a_n\}$各项均为正数，所以$a_n - a_{n - 1} = 1$，$n \geq 2$，所以$\{a_n\}$是公差为1的等差数列，所以$a_n = n$。
2. 由
(1)知$c_n = (-1)^n n(n + 1)$，当$n$为偶数时，$T_n = (-1) × 2 + 2 × 3 + (-3) × 4 + 4 × 5 + \cdots + [-(n - 1) × n] + n × (n + 1) = 2 × (-1 + 3) + 4 × (-3 + 5) + \cdots + n × [-(n - 1) + n + 1] = 2 × (2 + 4 + \cdots + n) = 2 × \frac{\frac{n}{2}(2 + n)}{2} = \frac{n(n + 2)}{2}$；当$n(n \geq 3)$为奇数时，$T_n = T_{n - 1} - n(n + 1) = \frac{(n - 1)(n + 1)}{2} - n(n + 1) = -\frac{(n + 1)^2}{2}$，经检验，$n = 1$也满足上式。

例 2 (2023·新高考Ⅱ卷节选)已知$a_{n}=2n + 3$，$b_{n}=\begin{cases}a_{n}-6,n 是奇数,\\2a_{n},n 是偶数.\end{cases}$记$S_{n}$，$T_{n}$分别为数列$\{ a_{n}\}$，$\{ b_{n}\}$的前$n$项和，证明：当$n ＞ 5$时，$T_{n}$＞$＞ S_{n}$。

答案： 因为$a_n = \begin{cases} 2n - 3, & n 是奇数, \\ 4n + 6, & n 是偶数, \end{cases}$所以$S_n = \frac{n[5 + (2n + 3)]}{2} = n^2 + 4n$。当$n$为奇数时，$T_n = (-1 + 14) + (3 + 22) + (7 + 30) + \cdots + [(2n - 7) + (2n - 3)] + [14 + 22 + 30 + \cdots + (4n + 2)] = \frac{n + 1}{2}(-1 + 2n - 3) + \frac{n - 1}{2}(14 + 4n + 2) = \frac{3n^2 + 5n - 10}{2}$。当$n ＞ 5$时，$T_n - S_n = \frac{3n^2 + 5n - 10}{2} - (n^2 + 4n) = \frac{n^2 - 3n - 10}{2} = \frac{(n - 5)(n + 2)}{2} ＞ 0$，所以$T_n ＞ S_n$；当$n$为偶数时，$T_n = (-1 + 14) + (3 + 22) + (7 + 30) + \cdots + [(2n - 5) + (4n + 6)] = [-1 + 3 + 7 + \cdots + (2n - 5)] + [14 + 22 + 30 + \cdots + (4n + 6)] = \frac{\frac{n}{2}(-1 + 2n - 5)}{2} + \frac{\frac{n}{2}(14 + 4n + 6)}{2} = \frac{3n^2 + 7n}{2}$。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，$T_n ＞ S_n$。

已知数列$\{ a_{n}\}$的前$n$项和为$S_{n}$，满足$a_{1}=2$，且$a_{n}$是$2$与$S_{n}$的等差中项。
(1)求数列$\{ a_{n}\}$的通项公式；
(2)若$b_{n}=(-1)^{n}\cdot\log_{2}a_{2n + 1}$，求数列$\{ b_{n}\}$的前$n$项和$T_{n}$。解 (1)因为$a_n$是2与$S_n$的等差中项，所以$2a_n = S_n + 2$，当$n \geq 2$时，$2a_{n - 1} = S_{n - 1} + 2$，① - ②得$2a_n - 2a_{n - 1} = a_n$，所以$a_n = 2a_{n - 1}(n \geq 2)$，又$a_1 = 2$，所以$\{a_n\}$是以2为首项，2为公比的等比数列，所以$a_n = 2^n$。(2)$b_n = (-1)^n \cdot \log_2 a_{2n + 1} = (-1)^n \cdot (2n + 1)$，当$n$为偶数时，$T_n = (b_1 + b_2) + (b_3 + b_4) + \cdots + (b_{n - 1} + b_n) = (-3 + 5) + (-7 + 9) + \cdots + [-(2n - 1) + (2n + 1)] = 2 + 2 + \cdots + 2 = 2 \cdot \frac{n}{2} = n$；当$n$为奇数时，$T_n = b_1 + (b_2 + b_3) + (b_4 + b_5) + \cdots + (b_{n - 1} + b_n) = -3 + (-2) + (-2) + \cdots + (-2) = -3 + (-2) \cdot \frac{n - 1}{2} = -n - 2$。综上所述，数列$\{b_n\}$的前$n$项和$T_n = \begin{cases} -n - 2, & n 为奇数, \\ n, & n 为偶数. \end{cases}$

答案： 1. 因为$a_n$是2与$S_n$的等差中项，所以$2a_n = S_n + 2$，当$n \geq 2$时，$2a_{n - 1} = S_{n - 1} + 2$，① - ②得$2a_n - 2a_{n - 1} = a_n$，所以$a_n = 2a_{n - 1}(n \geq 2)$，又$a_1 = 2$，所以$\{a_n\}$是以2为首项，2为公比的等比数列，所以$a_n = 2^n$。
2. $b_n = (-1)^n \cdot \log_2 a_{2n + 1} = (-1)^n \cdot (2n + 1)$，当$n$为偶数时，$T_n = (b_1 + b_2) + (b_3 + b_4) + \cdots + (b_{n - 1} + b_n) = (-3 + 5) + (-7 + 9) + \cdots + [-(2n - 1) + (2n + 1)] = 2 + 2 + \cdots + 2 = 2 \cdot \frac{n}{2} = n$；当$n$为奇数时，$T_n = b_1 + (b_2 + b_3) + (b_4 + b_5) + \cdots + (b_{n - 1} + b_n) = -3 + (-2) + (-2) + \cdots + (-2) = -3 + (-2) \cdot \frac{n - 1}{2} = -n - 2$。综上所述，数列$\{b_n\}$的前$n$项和$T_n = \begin{cases} -n - 2, & n 为奇数, \\ n, & n 为偶数. \end{cases}$