《2025年创新设计高考总复习数学浙江专版》

例 3 (2025·丽水、湖州、衢州模拟)设等差数列$\{ a_{n}\}$的公差为$d$，记$S_{n}$是数列$\{ a_{n}\}$的前$n$项和，若$S_{5}=a_{3}+20$，$S_{15}=a_{2}a_{3}a_{8}$。
(1)求数列$\{ a_{n}\}$的通项公式；
(2)若$d ＞ 0$，$b_{n}=\frac{4S_{n}}{a_{n}\cdot a_{n + 1}}(n\in\mathbf{N}^{*})$，数列$\{ b_{n}\}$的前$n$项和为$T_{n}$，求证：$T_{n} ＜ n+\frac{1}{2}$。解 (1)由$S_3 = a_3 + 20$得$5a_3 = a_3 + 20$，故$a_3 = 5$。由$S_{15} = a_2 a_3 a_8$得$15a_8 = a_2 a_3 a_8 = 5a_2 a_8$，所以$a_8 = 0$或$a_2 = 3$，当$a_8 = 0$时，$d = \frac{a_8 - a_3}{5} = -1$，此时$a_n = a_3 + (n - 3) × d = 8 - n$；当$a_2 = 3$时，$d = a_3 - a_2 = 2$，此时$a_n = a_3 + (n - 3) × d = 2n - 1$。综上，数列$\{a_n\}$的通项公式为$a_n = 8 - n$或$a_n = 2n - 1$。

答案： 1. 由$S_3 = a_3 + 20$得$5a_3 = a_3 + 20$，故$a_3 = 5$。由$S_{15} = a_2 a_3 a_8$得$15a_8 = a_2 a_3 a_8 = 5a_2 a_8$，所以$a_8 = 0$或$a_2 = 3$，当$a_8 = 0$时，$d = \frac{a_8 - a_3}{5} = -1$，此时$a_n = a_3 + (n - 3) × d = 8 - n$；当$a_2 = 3$时，$d = a_3 - a_2 = 2$，此时$a_n = a_3 + (n - 3) × d = 2n - 1$。综上，数列$\{a_n\}$的通项公式为$a_n = 8 - n$或$a_n = 2n - 1$。
2. 证明 因为$d ＞ 0$，所以$a_n = 2n - 1$，所以$S_n = n^2$。则$b_n = \frac{4S_n}{a_n \cdot a_{n + 1}} = \frac{4n^2}{(2n - 1)(2n + 1)} = \frac{4n^2 - 1 + 1}{(2n - 1)(2n + 1)} = 1 + \frac{1}{2}(\frac{1}{2n - 1} - \frac{1}{2n + 1})$。所以$T_n = b_1 + b_2 + \cdots + b_n = 1 + \frac{1}{2}(1 - \frac{1}{3}) + 1 + \frac{1}{2}(\frac{1}{3} - \frac{1}{5}) + \cdots + 1 + \frac{1}{2}(\frac{1}{2n - 1} - \frac{1}{2n + 1}) = n + \frac{1}{2}(1 - \frac{1}{2n + 1}) = n + \frac{n}{2n + 1} = n + \frac{1}{2(2n + 1)} ＜ n + \frac{1}{2}$。

例 4 (2025·重庆部分学校联考)已知数列$\{ a_{n}\}$满足$a_{1}=1$，$a_{n}+a_{n + 1}=2^{n + 1}$。
(1)证明：数列$\left\{\frac{a_{n}}{2^{n}}-\frac{2}{3}\right\}$为等比数列，并求数列$\{ a_{n}\}$的通项公式；
(2)设$T_{n}=\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{n}}$，证明：$T_{n} ＜ \frac{7}{4}$对任意$n\in\mathbf{N}^{*}$恒成立。证明 (1)由$a_{n + 1} + a_n = 2^{n + 1}$得$\frac{a_{n + 1}}{2^{n + 1}} + \frac{1}{2} × \frac{a_n}{2^n} = 1$，因此$\frac{a_{n + 1}}{2^{n + 1}} - \frac{2}{3} = -\frac{1}{2}(\frac{a_n}{2^n} - \frac{2}{3})$，又$\frac{a_1}{2^1} - \frac{2}{3} = \frac{1}{6}$，所以数列$\{\frac{a_n}{2^n} - \frac{2}{3}\}$是首项为$-\frac{1}{6}$，公比为$-\frac{1}{2}$的等比数列，因此，$\frac{a_n}{2^n} - \frac{2}{3} = -\frac{1}{6} × (-\frac{1}{2})^{n - 1}$，所以$a_n = \frac{2^{n + 1} + (-1)^n}{3}$。(2)由已知得$a_1 = 1$，$a_2 = 3$，$a_3 = 5$，$\cdots$，显然$T_n$单调递增，$T_1 ＜ T_2 = \frac{1}{a_1} + \frac{1}{a_2} = \frac{4}{7} ＜ \frac{7}{4}$。当$n ＞ 2$且$n$是奇数时，$\frac{1}{a_{n - 1}} + \frac{1}{a_n} = 3 × \frac{2^n + 2^{n + 1}}{(2^n + 1)(2^{n + 1} + 1)} ＜ 3 × \frac{2^n + 2^{n + 1}}{2^n × 2^{n + 1}} = 3(\frac{1}{2^n} + \frac{1}{2^{n + 1}})$，所以$T_n = \frac{1}{a_1} + \frac{1}{a_2} + \cdots + \frac{1}{a_n} ＜ 1 + 3 × [(\frac{1}{2^3} + \frac{1}{2^4}) + \cdots + (\frac{1}{2^n} + \frac{1}{2^{n + 1}})] = 1 + 3 × (\frac{1}{4} - \frac{1}{2^{n + 1}}) = \frac{7}{4} - \frac{3}{2^{n + 1}} ＜ \frac{7}{4}$。当$n ＞ 2$且$n$是偶数时，$n + 1$是奇数，有$T_n ＜ T_{n + 1} ＜ \frac{7}{4}$。所以对任意$n \in N^*$，$T_n = \frac{1}{a_1} + \frac{1}{a_2} + \cdots + \frac{1}{a_n} ＜ \frac{7}{4}$。

答案： 1. 证明
(1)由$a_{n + 1} + a_n = 2^{n + 1}$得$\frac{a_{n + 1}}{2^{n + 1}} + \frac{1}{2} × \frac{a_n}{2^n} = 1$，因此$\frac{a_{n + 1}}{2^{n + 1}} - \frac{2}{3} = -\frac{1}{2}(\frac{a_n}{2^n} - \frac{2}{3})$，又$\frac{a_1}{2^1} - \frac{2}{3} = \frac{1}{6}$，所以数列$\{\frac{a_n}{2^n} - \frac{2}{3}\}$是首项为$-\frac{1}{6}$，公比为$-\frac{1}{2}$的等比数列，因此，$\frac{a_n}{2^n} - \frac{2}{3} = -\frac{1}{6} × (-\frac{1}{2})^{n - 1}$，所以$a_n = \frac{2^{n + 1} + (-1)^n}{3}$。
2. 由已知得$a_1 = 1$，$a_2 = 3$，$a_3 = 5$，$\cdots$，显然$T_n$单调递增，$T_1 ＜ T_2 = \frac{1}{a_1} + \frac{1}{a_2} = \frac{4}{7} ＜ \frac{7}{4}$。当$n ＞ 2$且$n$是奇数时，$\frac{1}{a_{n - 1}} + \frac{1}{a_n} = 3 × \frac{2^n + 2^{n + 1}}{(2^n + 1)(2^{n + 1} + 1)} ＜ 3 × \frac{2^n + 2^{n + 1}}{2^n × 2^{n + 1}} = 3(\frac{1}{2^n} + \frac{1}{2^{n + 1}})$，所以$T_n = \frac{1}{a_1} + \frac{1}{a_2} + \cdots + \frac{1}{a_n} ＜ 1 + 3 × [(\frac{1}{2^3} + \frac{1}{2^4}) + \cdots + (\frac{1}{2^n} + \frac{1}{2^{n + 1}})] = 1 + 3 × (\frac{1}{4} - \frac{1}{2^{n + 1}}) = \frac{7}{4} - \frac{3}{2^{n + 1}} ＜ \frac{7}{4}$。当$n ＞ 2$且$n$是偶数时，$n + 1$是奇数，有$T_n ＜ T_{n + 1} ＜ \frac{7}{4}$。所以对任意$n \in N^*$，$T_n = \frac{1}{a_1} + \frac{1}{a_2} + \cdots + \frac{1}{a_n} ＜ \frac{7}{4}$。

已知函数$f(x)=2\sin\frac{\pi}{2}x$，把方程$\vert f(x)\vert = 2$的正数解从小到大依次排成一列，得到数列$\{ a_{n}\}$，$n\in\mathbf{N}^{*}$。
(1)求数列$\{ a_{n}\}$的通项公式；
(2)记$b_{n}=\frac{1}{a_{n + 1}^{2}}$，设数列$\{ b_{n}\}$的前$n$项和为$T_{n}$，求证：$T_{n} ＜ \frac{1}{4}$。解 (1)因为$f(x) = 2\sin \frac{\pi}{2}x$，令$|f(x)| = 2$，即$|\sin \frac{\pi}{2}x| = 1$，所以$\frac{\pi}{2}x = k\pi + \frac{\pi}{2}$，$k \in Z$，解得$x = 2k + 1(k \in Z)$，所以方程$|f(x)| = 2$的正数解从小到大依次为$1,3,5,7,\cdots$，所以$a_n = 2n - 1$。(2)证明 记$b_n = \frac{1}{a_n^2 a_{n + 1}^2}$，数列$\{b_n\}$的前$n$项和为$T_n$，$b_n = \frac{1}{a_n^2 a_{n + 1}^2} = \frac{1}{(2n - 1)^2(2n + 1)^2} = \frac{1}{4n^2 + 4n + 1} ＜ \frac{1}{4n^2 + 4n} = \frac{1}{4}(\frac{1}{n} - \frac{1}{n + 1})$，所以$T_n = b_1 + b_2 + \cdots + b_n ＜ \frac{1}{4}(1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \cdots + \frac{1}{n} - \frac{1}{n + 1}) = \frac{1}{4}(1 - \frac{1}{n + 1}) ＜ \frac{1}{4}$。

答案： 1. 解
(1)因为$f(x) = 2\sin \frac{\pi}{2}x$，令$|f(x)| = 2$，即$|\sin \frac{\pi}{2}x| = 1$，所以$\frac{\pi}{2}x = k\pi + \frac{\pi}{2}$，$k \in Z$，解得$x = 2k + 1(k \in Z)$，所以方程$|f(x)| = 2$的正数解从小到大依次为$1,3,5,7,\cdots$，所以$a_n = 2n - 1$。
2. 证明 记$b_n = \frac{1}{a_n^2 a_{n + 1}^2}$，数列$\{b_n\}$的前$n$项和为$T_n$，$b_n = \frac{1}{a_n^2 a_{n + 1}^2} = \frac{1}{(2n - 1)^2(2n + 1)^2} = \frac{1}{4n^2 + 4n + 1} ＜ \frac{1}{4n^2 + 4n} = \frac{1}{4}(\frac{1}{n} - \frac{1}{n + 1})$，所以$T_n = b_1 + b_2 + \cdots + b_n ＜ \frac{1}{4}(1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \cdots + \frac{1}{n} - \frac{1}{n + 1}) = \frac{1}{4}(1 - \frac{1}{n + 1}) ＜ \frac{1}{4}$。