《2025年学霸高考黑题数学人教版》

变式训练2. (2025·山东聊城期中)设$x\in\mathbf{R}$，用$[x]$表示不超过$x$的最大整数.已知数列$\{a_n\}$满足$a_2 = 1$，$2S_n = na_n$，若$b_n = [\lg(a_n + 1)]$，数列$\{b_n\}$的前$n$项和为$T_n$，则$T_{2024}=$ (B)

A.$4956$
B.$4965$
C.$7000$
D.$8022$

答案： 变式训练2. B 解析：当$n = 1$时，$2S_1 = 1 × a_1$. 因为$S_1 = a_1$，所以$a_1 = 0$. 当$n \geq 2$时，$2S_n = na_n$，$2S_{n - 1} = (n - 1)a_{n - 1}$. 两式相减得$2a_n = na_n - (n - 1)a_{n - 1}$，即$(n - 2)a_n = (n - 1)a_{n - 1}$. 当$n \geq 3$时，那么$\frac{a_n}{a_{n - 1}} = \frac{n - 1}{n - 2}$，那么$a_n = a_2 × \frac{a_3}{a_2} × \frac{a_4}{a_3} × ·s × \frac{a_n}{a_{n - 1}} = 1 × \frac{2}{1} × \frac{3}{2} × ·s × \frac{n - 1}{n - 2} = n - 1(n \geq 3)$. 当$n = 1$时，$a_1 = 0$也满足$a_n = n - 1$. 当$n = 2$时，$a_2 = 1$，当$n = 1$时，$a_1 = 0$，$b_1 = [\lg(a_1 + 1)] = [\lg 1] = 0$；$·s$；当$n = 10$时，$a_{10} = 9$，$b_{10} = [\lg(a_{10} + 1)] = [\lg 10] = 1$. 可以发现当$1 \leq n \leq 9$时，$b_n = 0$，共$9$个. 当$10 \leq n \leq 99$时，$1 \leq \lg(n + 1) ＜ 2$，所以$b_n = 1$，共有$99 - 10 + 1 = 90$(个). $1 \leq \lg(n - 1 + 1) = \lg n ＜ 3$，所以$b_n = 2$，共有$999 - 100 + 1 = 900$(个). 当$1000 \leq n \leq 2024$时，$3 \leq \lg(n - 1 + 1) = \lg n ＜ 4$，所以$b_n = 3$，共有$2024 - 1000 + 1 = 1025$(个). 则$T_{2024} = 9 × 0 + 90 × 1 + 900 × 2 + 1025 × 3 = 0 + 90 + 1800 + 3075 = 4965$.

变式训练3. (2025·广东江门期中)在数列$\{a_n\}$中，$a_1 = 2$，$a_2 = 8$，且对任意的$n\in\mathbf{N}^*$，都有$a_{n + 2}=4a_{n + 1}-4a_n$，则$\{a_n\}$的通项公式为$a_n = n · 2^n$；若$b_n=\begin{cases}\frac{n}{a_n},n = 2k - 1,k\in\mathbf{N}^*,\\\log_2\frac{n}{a_n},n = 2k,k\in\mathbf{N}^*,\end{cases}$则数列$\{b_n\}$的前$n$项和$T_n=$$\frac{1}{3 × 2^n} - \frac{11}{12}$，$n = 2k - 1, k \in \mathbf{N}^*$，$\frac{1}{3 × 2^{n - 1}} - \frac{2}{3}$，$n = 2k, k \in \mathbf{N}^*$.

答案： 变式训练3. $a_n = n · 2^n$ $\frac{1}{3 × 2^n} - \frac{11}{12}$，$n = 2k - 1, k \in \mathbf{N}^*$，$\frac{1}{3 × 2^{n - 1}} - \frac{2}{3}$，$n = 2k, k \in \mathbf{N}^*$
解析：因为$a_1 = 2$，$a_2 = 8$，所以$a_2 - 2a_1 = 8 - 2 × 2 = 4$. 因为$a_{n + 2} = 4a_{n + 1} - 4a_n$，所以$a_{n + 2} - 2a_{n + 1} = 2(a_{n + 1} - 2a_n)$. 又$a_2 - 2a_1 = 4 \neq 0$，则有$a_{n + 1} - 2a_n \neq 0(n \in \mathbf{N}^*)$，所以$\frac{a_{n + 2} - 2a_{n + 1}}{a_{n + 1} - 2a_n} = 2$，所以$\{ a_{n + 1} - 2a_n \}$是以$4$为首项，$2$为公比的等比数列. 所以$a_{n + 1} - 2a_n = 4 × 2^{n - 1} = 2^{n + 1}$，所以$\frac{a_{n + 1}}{2^{n + 1}} - \frac{a_n}{2^n} = 1$. 又$\frac{a_1}{2} = 1$，所以$\{ \frac{a_n}{2^n} \}$是以$1$为首项，$1$为公差的等差数列，所以$\frac{a_n}{2^n} = 1 + (n - 1) × 1 = n$，所以$a_n = n · 2^n$.
由题意可得$b_n = \begin{cases} \frac{n}{2^n}, n = 2k - 1, k \in \mathbf{N}^*, \\ \frac{1}{2^n}, n = 2k - 1, k \in \mathbf{N}^*, \end{cases}$则$\{ b_n \}$的奇数项为以$b_1 = \frac{1}{2}$为首项，$\frac{1}{4}$为公比的等比数列；偶数项是以$b_2 = -2$为首项，$-2$为公差的等差数列. 所以当$n$为偶数，且$n \geq 2$时，$T_n = (b_1 + b_3 + ·s + b_{n - 1}) + (b_2 + b_4 + ·s + b_n) = \left( \frac{\frac{1}{2} × \left[ 1 - \left( \frac{1}{4} \right)^{\frac{n}{2}} \right]}{1 - \frac{1}{4}} \right) + \frac{n(-2 - n)}{2} = \frac{2}{3} × \left( 1 - \frac{1}{2^n} \right) - \frac{n(n + 2)}{4}$. 当$n$为奇数，且$n \geq 3$时，$n - 1$为偶数，$T_n = T_{n - 1} + b_n = \frac{1}{3 × 2^{n - 2}} × \frac{n - 1}{2} × \frac{2}{3} - \frac{1}{2^n} = \frac{1}{3 × 2^{n - 1}} × \frac{n^2}{4} - \frac{11}{12}$. 当$n = 1$时，$T_1 = \frac{1}{3 × 2^1} + \frac{11}{12} = \frac{1}{6} + \frac{11}{12} = \frac{13}{12}$，满足. 所以，当$n$为奇数，$n \geq 1$时，$T_n = \frac{1}{3 × 2^{n - 1}} + \frac{n^2}{4} - \frac{11}{12}$，$n = 2k, k \in \mathbf{N}^*$.

典型例题3. (2022·新高考全国Ⅰ)记$S_n$为数列$\{a_n\}$的前$n$项和，已知$a_1 = 1$，$\{\frac{S_n}{a_n}\}$是公差为$\frac{1}{3}$的等差数列.
(1)求$\{a_n\}$的通项公式；
(2)证明：$\frac{1}{a_1}+\frac{1}{a_2}+·s+\frac{1}{a_n}＜2$.

答案： 典型例题3.
(1) 解：$\because a_1 = 1$，$\therefore S_1 = a_1 = 1$，$\therefore \frac{S_1}{a_1} = 1$. 又$\{ \frac{S_n}{a_n} \}$为公差为$\frac{1}{3}$的等差数列，$\therefore \frac{S_n}{a_n} = 1 + \frac{1}{3}(n - 1) = \frac{n + 2}{3}$，$\therefore S_n = \frac{(n + 2)a_n}{3}$，当$n \geq 2$时，$S_{n - 1} = \frac{(n + 1)a_{n - 1}}{3}$，$\therefore a_n = S_n - S_{n - 1} = \frac{(n + 2)a_n}{3} - \frac{(n + 1)a_{n - 1}}{3}$，整理得$(n - 1)a_n = (n + 1)a_{n - 1}$，即$\frac{a_n}{a_{n - 1}} = \frac{n + 1}{n - 1}$，$\therefore a_n = a_1 × \frac{a_2}{a_1} × \frac{a_3}{a_2} × ·s × \frac{a_{n - 1}}{a_{n - 2}} × \frac{a_n}{a_{n - 1}} = 1 × \frac{3}{1} × \frac{4}{2} × ·s × \frac{n}{n - 2} × \frac{n + 1}{n - 1} = \frac{n(n + 1)}{2}(n \geq 2)$，显然对于$n = 1$也成立，$\therefore \{ a_n \}$的通项公式为$a_n = \frac{n(n + 1)}{2}(n \in \mathbf{N}^*)$.
(2) 证明：$\because \frac{1}{a_n} = \frac{2}{n(n + 1)} = 2 \left( \frac{1}{n} - \frac{1}{n + 1} \right)$，$\therefore \frac{1}{a_1} + \frac{1}{a_2} + ·s + \frac{1}{a_n} = 2 \left[ \left( \frac{1}{1} - \frac{1}{2} \right) + \left( \frac{1}{2} - \frac{1}{3} \right) + ·s + \left( \frac{1}{n} - \frac{1}{n + 1} \right) \right] = 2 \left( 1 - \frac{1}{n + 1} \right) ＜ 2$.

变式训练4. (2025·福建龙岩二模)已知数列$\{a_n\}$的前$n$项和为$S_n$，且满足$nS_{n + 1}-(n + 1)S_n = n(n + 1)$，$n\in\mathbf{N}^*$，$a_1 = 1$.
(1)求数列$\{a_n\}$的通项公式；
(2)若$b_n = (-1)^n·\frac{2a_n + 2}{a_na_{n + 1}}$，求数列$\{b_n\}$的前$n$项和$T_n$.

答案： 变式训练4. 解：
(1) 由$nS_{n + 1} - (n + 1)S_n = n(n + 1), n \in \mathbf{N}^*$，得$\frac{S_{n + 1}}{n + 1} - \frac{S_n}{n} = 1$. 又$a_1 = 1$，$\therefore$数列$\{ \frac{S_n}{n} \}$是首项为$\frac{S_1}{1} = a_1 = 1$，公差为$d = 1$的等差数列，$\therefore \frac{S_n}{n} = 1 + (n - 1) × 1 = n$，即$S_n = n^2$，当$n \geq 2$时，$a_n = S_n - S_{n - 1} = n^2 - (n - 1)^2 = 2n - 1$，且$a_1 = 1$也满足，$\therefore a_n = 2n - 1$，则数列$\{ a_n \}$的通项公式为$a_n = 2n - 1$.
(2) 由
(1) 得$a_n = 2n - 1$，$\therefore b_n = (-1)^n · \frac{4n}{(2n - 1)(2n + 1)} = (-1)^n · \left( \frac{1}{2n - 1} + \frac{1}{2n + 1} \right)$，$\therefore T_n = - \left( 1 + \frac{1}{3} \right) + \left( \frac{1}{3} + \frac{1}{5} \right) - \left( \frac{1}{5} + \frac{1}{7} \right) + ·s + (-1)^n · \left( \frac{1}{2n - 1} + \frac{1}{2n + 1} \right) = -1 + (-1)^n · \frac{1}{2n + 1}$.