《2026年南方凤凰台5A新考案高中数学二轮基础版》

例 1 (1)已知数列$\{ a_{n}\}$满足$a_{n}a_{n + 4}=-\frac{1}{4}$，$a_{4}=\frac{1}{2}$，则$a_{200}=$(B)

A.$\frac{1}{2}$
B.$-\frac{1}{2}$
C.$0$
D.$-\frac{1}{4}$

答案： 例1
(1) B 【解析】由$a_{n}a_{n+4} = - \frac{1}{4}，$得$a_{n} \neq 0$且$a_{n+4}a_{n+8} = - \frac{1}{4}，$所以$a_{n+8}a_{n+4} = a_{n}a_{n+4}，$故$a_{n+8} = a_{n}，$所以$\{a_{n}\}$是以8为一个周期的周期数列.因为$a_{4} = \frac{1}{2}，$$a_{4}a_{8} = - \frac{1}{4}，$所以$a_{8} = - \frac{1}{2}，$所以$a_{200} = a_{25 × 8} = a_{8} = - \frac{1}{2}.$

(2)(2025·湖北二模)若数列$\{ a_{n}\}$满足$a_{1}=2$，$a_{n + 1}=\frac{1 + a_{n}}{1 - a_{n}}(n\in \mathbf{N}^{*})$，则该数列的前$2025$项的乘积是(C)

A.$-2$
B.$-1$
C.$2$
D.$1$

答案： 例1
(2) C 【解析】因为数列$\{a_{n}\}$满足$a_{1} = 2，$$a_{n+1} = \frac{1 + a_{n}}{1 - a_{n}}(n \in \mathbf{N}^*)，$所以$a_{2} = \frac{1 + a_{1}}{1 - a_{1}} = \frac{1 + 2}{1 - 2} = - 3，$同理可得$a_{3} = - \frac{1}{2}，$$a_{4} = \frac{1}{3}，$$a_{5} = 2，$·s，所以数列$\{a_{n}\}$的周期为4，即$a_{n+4} = a_{n}，$且$a_{1} · a_{2} · a_{3} · a_{4} = 1.$而2025 = 506 × 4 + 1，所以该数列的前2025项的乘积是$a_{1} · a_{2} · a_{3} · a_{4} · ·s · a_{2025} = 1^{506} × a_{1} = 2.$

变式 1 (2025·湖北一模)
已知数列$\{ a_{n}\}$满足：①任意相邻两项的乘积不等于$1$；②任意连续三项的乘积等于这三项的和；③$a_{1}=2$，$a_{2}=3$。记数列$\{ a_{n}\}$的前$n$项和为$S_{n}$，则$S_{2025}-S_{2012}$的值为(C)

A.$27$
B.$26$
C.$25$
D.$24$

答案： 变式1 C 【解析】依题意$a_{n}a_{n+1} \neq 1$且$a_{n}a_{n+1}a_{n+2} = a_{n} + a_{n+1} + a_{n+2}，$则$a_{n+1}a_{n+2}a_{n+3} = a_{n+1} + a_{n+2} + a_{n+3}，$两式相减得$a_{n+1}a_{n+2}(a_{n} - a_{n+3}) = a_{n} - a_{n+3}，$故$(a_{n+1}a_{n+2} - 1)(a_{n} - a_{n+3}) = 0.$因为$a_{n+1}a_{n+2} \neq 1，$所以$a_{n} - a_{n+3} = 0，$故$a_{n} = a_{n+3}，$故数列$\{a_{n}\}$是周期为3的数列.由$a_{1} = 2，$$a_{2} = 3$及$a_{1}a_{2}a_{3} = a_{1} + a_{2} + a_{3}，$可得$a_{3} = 1，$所以$S_{2025} - S_{2012} = a_{2013} + a_{2014} + ·s + a_{2025} = (a_{1} + a_{2} + a_{3}) × 4 + a_{3} = (2 + 3 + 1) × 4 + 1 = 25.$

例 2 (1)(2025·重庆期初)已知等差数列$\{ a_{n}\}$的公差$d\lt0$，$a_{5}a_{7}=35$，$a_{4}+a_{8}=12$，记该数列的前$n$项和为$S_{n}$。若数列$\{ S_{n}+\lambda n\}$是递减数列，则实数$\lambda$的取值范围是(D)

A.$(-\infty,-\frac{21}{2}]$
B.$(-\infty,-\frac{21}{2})$
C.$(-\infty,-10]$
D.$(-\infty,-10)$

答案： 例2
(1) D 【解析】在等差数列$\{a_{n}\}$中，$a_{5} + a_{7} = a_{4} + a_{8} = 12，$$a_{5}a_{7} = 35，$由公差d ＜ 0，得a_{5} ＞$ a_{7}，$解得$a_{5} = 7，$$a_{7} = 5，$$d = \frac{a_{7} - a_{5}}{7 - 5} = - 1，$所以$a_{n} = a_{5} + (n - 5)d = - n + 12，$$S_{n} = \frac{11 + (-n + 12)}{2} · n = - \frac{1}{2}n^{2} + \frac{23}{2}n，$$S_{n} + \lambda n = - \frac{1}{2}n^{2} + (\lambda + \frac{23}{2})n.$由数列$\{S_{n} + \lambda n\}$是递减数列，得$ - \frac{1}{2}(n + 1)^{2} + (\lambda + \frac{23}{2})(n + 1) ＜ - \frac{1}{2}n^{2} + (\lambda + \frac{23}{2})n$恒成立，整理得$\lambda ＜ n -$
值范围是$(- \infty, - 10).$

(2)已知数列$\{ a_{n}\}$的前$n$项和为$S_{n}$，$a_{1}=1$，$a_{n + 1}-S_{n}=1$，$b_{n}=\frac{4n^{2}}{a_{n}}$，则$b_{n}$的最大值为(B)

A.$4$
B.$9$
C.$10$
D.$12$

答案： 例2
(2) B 【解析】当n = 1时，$a_{2} = 1 + S_{1} = 1 + a_{1} = 2;$当$n \geq 2$时，$a_{n} = 1 + S_{n - 1}，$所以$a_{n + 1} - a_{n} = (1 + S_{n}) - (1 + S_{n - 1}) = S_{n} - S_{n - 1} = a_{n}，$即$a_{n + 1} = 2a_{n}.$又$a_{2} = 2a_{1}，$所以数列$\{a_{n}\}$是以1为首项，2为公比的等比数列，所以$a_{n} = 2^{n - 1}，$所以$b_{n} = \frac{4n^{2}}{2^{n - 1}}.$令$b_{n + 1} - b_{n} ＞ 0，$得$\frac{4(n + 1)^{2}}{2^{n}} ＞ \frac{4n^{2}}{2^{n - 1}}，$即$2n + 1 ＞ n^{2}，$则n = 1或n = 2.所以在数列$\{b_{n}\}$中，$b_{1} $＜ b_{2} ＜ b_{3} ＞$ b_{4} ＞ ·s，$即数列$\{b_{n}\}$中，$b_{3}$最大，且$b_{3} = \frac{4 × 3^{2}}{2^{2}} = 9.$