《2025年学霸高考黑题数学人教版》

典型例题1. (2025·浙江湖州一模)若$f(x)=(x - 2)^3 + 2(x - 2) + 2$，已知数列$\{a_n\}$中，首项$a_1=\frac{1}{20}$，$a_n = a_1 + \frac{a_2}{2} + \frac{a_3}{3} + ·s + \frac{a_n}{n}$，$n\in\mathbf{N}^*$，则$\sum_{i = 1}^{79}f(a_i)=$158.

答案： 典型例题1.158 解析：因为$a_n = a_1 + \frac{a_2}{2} + \frac{a_3}{3} + ·s + \frac{a_n}{n}, n \in \mathbf{N}^*$，所以$a_{n+1} = a_1 + \frac{a_2}{2} + \frac{a_3}{3} + ·s + \frac{a_n}{n} + \frac{a_{n+1}}{n+1}$. 所以$a_{n+1} = a_n + \frac{a_{n+1}}{n+1}$ 整理得$\frac{a_{n+1}}{n+1} = \frac{a_n}{n}$.
$n \in \mathbf{N}^*$，即$\left\{ \frac{a_n}{n} \right\}$是常数数列. 又$a_1 = \frac{1}{20}$，所以$\frac{a_n}{n} = \frac{a_1}{1} = \frac{1}{20}$，即$a_n = \frac{n}{20}$.
因为$f(x) = (x - 2)^3 + 2(x - 2) + 2$，所以$f(4 - x) = (4 - x - 2)^3 + 2(4 - x - 2) + 2 = -(x - 2)^3 - 2(x - 2) + 2$，所以$f(x) + f(4 - x) = 4$. 又$a_n = \frac{n}{20}$，所以$f(a_i) = f\left( \frac{i}{20} \right), i \in \mathbf{N}^*$，所以$\sum_{i=1}^{79} f(a_i) = f\left( \frac{1}{20} \right) + f\left( \frac{2}{20} \right) + ·s + f\left( \frac{78}{20} \right) + f\left( \frac{79}{20} \right)$，即$\sum_{i=1}^{79} f(a_i) = f\left( \frac{79}{20} \right) + f\left( \frac{78}{20} \right) + ·s + f\left( \frac{2}{20} \right) + f\left( \frac{1}{20} \right)$，所以$2 \sum_{i=1}^{79} f(a_i) = \left( f\left( \frac{1}{20} \right) + f\left( \frac{79}{20} \right) \right) + \left( f\left( \frac{2}{20} \right) + f\left( \frac{78}{20} \right) \right) + ·s + \left( f\left( \frac{79}{20} \right) + f\left( \frac{1}{20} \right) \right) = 4 × 79$，所以$\sum_{i=1}^{79} f(a_i) = 2 × 79 = 158$.

变式训练1. (2025·江西南昌期中)已知函数$f(x)$满足$f(x)=f(1 - x)$，$f'(x)$为$f(x)$的导函数，$g(x)=f'(x)+\frac{1}{3}$，$x\in\mathbf{R}$.若$a_n = g(\frac{n}{2026})$，则数列$\{a_n\}$的前$2025$项和为675.

答案： 变式训练1.675 解析：由题意知$f(x) = f(1 - x)$，所以$f'(x) = -f'(1 - x)$，即$f'(x) + f'(1 - x) = 0$. 又因为$g(x) = f'(x) + \frac{1}{3}$，所以$g(x) + g(1 - x) = f'(x) + f'(1 - x) + \frac{2}{3} = \frac{2}{3}$，所以$a_1 + a_2 + a_3 + ·s + a_{2025} = g\left( \frac{1}{2026} \right) + g\left( \frac{2}{2026} \right) + g\left( \frac{3}{2026} \right) + ·s + g\left( \frac{2025}{2026} \right)$ ①，$a_1 + a_2 + a_3 + ·s + a_{2025} = g\left( \frac{2025}{2026} \right) + g\left( \frac{2024}{2026} \right) + g\left( \frac{2023}{2026} \right) + ·s + g\left( \frac{1}{2026} \right)$ ②，将①②两式相加可得$a_1 + a_2 + a_3 + ·s + a_{2025} = \frac{2025 × \frac{2}{3}}{2} = \frac{2025}{3} = 675$.

典型例题2. (2023·新课标全国Ⅱ)已知$\{a_n\}$为等差数列，$b_n=\begin{cases}a_n - 6,n为奇数,\\2a_n,n为偶数,\end{cases}$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： 典型例题2.
(1) 解：设等差数列$\{ a_n \}$的公差为$d$，而$b_n = \begin{cases} a_n - 6, n 为奇数, \\ 2a_n, n 为偶数, \end{cases}$则$b_1 = a_1 - 6, b_2 = 2a_2 = 2a_1 + 2d, b_3 = a_3 - 6 = a_1 + 2d - 6$，于是$\begin{cases} S_4 = 4a_1 + 6d = 32, \\ T_4 = 4a_1 + 4d - 12 = 16, \end{cases}$解得$\begin{cases} a_1 = 5, \\ d = 2, \end{cases}$所以$a_n = a_1 + (n - 1)d = 2n + 3$，所以数列$\{ a_n \}$的通项公式是$a_n = 2n + 3$.
(2) 证明：方法一：由
(1) 知，$S_n = \frac{n(5 + 2n + 3)}{2} = n^2 + 4n$，$b_n = \begin{cases} 2n - 3, n 为奇数, \\ 4n + 6, n 为偶数, \end{cases}$当$n$为偶数时，$b_{n - 1} + b_n = 2(n - 1) - 3 + 4n + 6 = 6n + 1$，$T_n = \frac{13 + (6n + 1)}{2} × \frac{n}{2} = \frac{3}{2}n^2 + \frac{7}{2}n$. 当$n ＞ 5$时，$T_n - S_n = \left( \frac{3}{2}n^2 + \frac{7}{2}n \right) - (n^2 + 4n) = \frac{1}{2}n(n - 1) ＞ 0$，因此$T_n ＞ S_n$；当$n$为奇数时，$T_n = T_{n - 1} + b_n = \frac{3}{2}(n - 1)^2 + \frac{7}{2}(n - 1) - [4(n + 1) + 6] = \frac{3}{2}n^2 + \frac{5}{2}n - 5$，当$n ＞ 5$时，$T_n - S_n = \left( \frac{3}{2}n^2 + \frac{5}{2}n - 5 \right) - (n^2 + 4n) = \frac{1}{2}(n + 2)(n - 5) ＞ 0$，因此$T_n ＞ S_n$，所以当$n ＞ 5$时，$T_n ＞ S_n$.
方法二：由
(1) 知，$S_n = \frac{n(5 + 2n + 3)}{2} = n^2 + 4n$，$b_n = \begin{cases} 2n - 3, n 为奇数, \\ 4n + 6, n 为偶数, \end{cases}$当$n$为偶数时，$T_n = (b_1 + b_3 + ·s + b_{n - 1}) + (b_2 + b_4 + ·s + b_n) = \frac{-1 + 2(n - 1) - 3}{2} × \frac{n}{2} + \frac{14 + 4n + 6}{2} × \frac{n}{2} = \frac{3}{2}n^2 + \frac{7}{2}n$，当$n ＞ 5$时，$T_n - S_n = \left( \frac{3}{2}n^2 + \frac{7}{2}n \right) - (n^2 + 4n) = \frac{1}{2}n(n - 1) ＞ 0$，因此$T_n ＞ S_n$；当$n$为奇数时，若$n \geq 3$，则$T_n = (b_1 + b_3 + ·s + b_n) + (b_2 + b_4 + ·s + b_{n - 1}) = \frac{-1 + 2n - 3}{2} × \frac{n + 1}{2} + \frac{14 + 4(n - 1) + 6}{2} × \frac{n - 1}{2} = \frac{3}{2}n^2 + \frac{5}{2}n - 5$，当$n ＞ 5$时，$T_n - S_n = \left( \frac{3}{2}n^2 + \frac{5}{2}n - 5 \right) - (n^2 + 4n) = \frac{1}{2}(n + 2)(n - 5) ＞ 0$，因此$T_n ＞ S_n$，所以当$n ＞ 5$时，$T_n ＞ S_n$.