《2025年金版新学案高中数学选择性必修第二册北师大版》

真题 1
(2024·全国甲卷)记$S_n$为等差数列$\{a_n\}$的前$n$项和，已知$S_5 = S_{10}$，$a_5 = 1$，则$a_1 =$（ ）

A.$\frac{7}{2}$
B.$\frac{7}{3}$
C.$-\frac{1}{3}$
D.$-\frac{7}{11}$

答案： B

真题 2
(2024·新课标Ⅱ卷)记$S_n$为等差数列$\{a_n\}$的前$n$项和。若$a_3 + a_4 = 7$，$3a_2 + a_5 = 5$，则$S_{10} =$______。

答案： 95

真题 3
(2024·全国甲卷)已知等比数列$\{a_n\}$的前$n$项和为$S_n$，且$2S_n = 3a_{n + 1} - 3$。
(1)求$\{a_n\}$的通项公式；
(2)求数列$\{S_n\}$的前$n$项和。

答案： 解：
(1)因为$2S_{n} = 3a_{n + 1} - 3$，两式相减可得$2a_{n + 1} = 3a_{n + 2} - 3a_{n + 1}$，即$a_{n + 2} = \frac{5}{3}a_{n + 1}$，所以等比数列$\{ a_{n}\}$的公比为$\frac{5}{3}$。因为$2S_{1} = 2a_{1} = 3a_{2} - 3 = 5a_{1} - 3$，所以$a_{1} = 1$，故$a_{n} = (\frac{5}{3})^{n - 1}$。
(2)因为$2S_{n} = 3a_{n + 1} - 3$，所以$S_{n} = \frac{3}{2}(a_{n + 1} - 1) = \frac{3}{2}[(\frac{5}{3})^{n} - 1]$，设数列$\{ S_{n}\}$的前$n$项和为$T_{n}$，则$T_{n} = \frac{3}{2} × \frac{\frac{5}{3} - (\frac{5}{3})^{n + 1}}{1 - \frac{5}{3}} - \frac{3}{2}n = \frac{15}{4} × (\frac{5}{3})^{n} - \frac{15}{4} - \frac{3}{2}n$。

真题 4
(2022·新高考Ⅰ卷)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： 解：
(1)因为$a_{1} = 1$，所以$S_{1} = a_{1} = 1$，所以$\frac{S_{1}}{a_{1}} = 1$，又因为$\{ \frac{S_{n}}{a_{n}}\}$是公差为$\frac{1}{3}$的等差数列，所以$\frac{S_{n}}{a_{n}} = 1 + \frac{1}{3}(n - 1) = \frac{n + 2}{3}$，所以$S_{n} = \frac{(n + 2)a_{n}}{3}$，所以当$n \geqslant 2$时，$S_{n - 1} = \frac{(n + 1)a_{n - 1}}{3}$，所以$a_{n} = S_{n} - S_{n - 1} = \frac{(n + 2)a_{n}}{3} - \frac{(n + 1)a_{n - 1}}{3}$，整理得$(n - 1)a_{n} = (n + 1)a_{n - 1}$，即$\frac{a_{n}}{a_{n - 1}} = \frac{n + 1}{n - 1}$，所以$a_{n} = a_{1} × \frac{a_{2}}{a_{1}} × \frac{a_{3}}{a_{2}} × ·s × \frac{a_{n}}{a_{n - 1}} = 1 × \frac{3}{1} × \frac{4}{2} × ·s × \frac{n}{n - 2} × \frac{n + 1}{n - 1} = \frac{n(n + 1)}{2}(n \geqslant 2)$，显然对于$n = 1$也成立，所以$\{ a_{n}\}$的通项公式为$a_{n} = \frac{n(n + 1)}{2}(n \in N^{*})$。
(2)证明：$\frac{1}{a_{n}} = \frac{2}{n(n + 1)} = 2(\frac{1}{n} - \frac{1}{n + 1})$，所以$\frac{1}{a_{1}} + \frac{1}{a_{2}} + ·s + \frac{1}{a_{n}} = 2[(1 - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + ·s + (\frac{1}{n} - \frac{1}{n + 1})] = 2(1 - \frac{1}{n + 1}) ＜ 2$。

真题 5
(2023·新课标Ⅰ卷)设等差数列$\{a_n\}$的公差为$d$，且$d ＞ 1$。令$b_n = \frac{n^2 + n}{a_n}$，记$S_n$，$T_n$分别为数列$\{a_n\}$，$\{b_n\}$的前$n$项和。
(1)若$3a_2 = 3a_1 + a_3$，$S_3 + T_3 = 21$，求$\{a_n\}$的通项公式；
(2)若$\{b_n\}$为等差数列，且$S_{99} - T_{99} = 99$，求$d$。

答案： 解：
(1)因为$3a_{2} = 3a_{1} + a_{3}$，所以$3d = a_{1} + 2d$，解得$a_{1} = d$，所以$S_{3} = 3a_{2} = 3(a_{1} + d) = 6d$，又$T_{3} = b_{1} + b_{2} + b_{3} = \frac{2}{d} + \frac{6}{2d} + \frac{12}{3d} = \frac{9}{d}$，所以$S_{3} + T_{3} = 6d + \frac{9}{d} = 21$，即$2d^{2} - 7d + 3 = 0$，解得$d = 3$或$d = \frac{1}{2}$(舍去)，所以$a_{n} = a_{1} + (n - 1) · d = 3n$。
(2)因为$\{ b_{n}\}$为等差数列，所以$2b_{2} = b_{1} + b_{3}$，即$\frac{12}{a_{2}} = \frac{2}{a_{1}} + \frac{12}{a_{3}}$，所以$6(\frac{1}{a_{2}} - \frac{1}{a_{3}}) = \frac{6d}{a_{2}a_{3}} = \frac{1}{a_{1}}$，即$a_{1}^{2} - 3a_{1}d + 2d^{2} = 0$，解得$a_{1} = d$或$a_{1} = 2d$，因为$d ＞ 1$，所以$a_{n} ＞ 0$，又$S_{99} - T_{99} = 99$，由等差数列性质知，$99a_{50} - 99b_{50} = 99$，即$a_{50} - b_{50} = 1$，所以$a_{50}^{2} - a_{50} - 2550 = 0$，解得$a_{50} = 51$或$a_{50} = -50$(舍去)。当$a_{1} = 2d$时，$a_{50} = a_{1} + 49d = 51d = 51$，解得$d = 1$，与$d ＞ 1$矛盾，舍去；当$a_{1} = d$时，$a_{50} = a_{1} + 49d = 50d = 51$，解得$d = \frac{51}{50}$。综上，$d = \frac{51}{50}$。

真题 6
(2023·新课标Ⅱ卷)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： 解：
(1)设等差数列$\{ a_{n}\}$的公差为$d$，而$b_{n} = \begin{cases}a_{n} - 6,n = 2k - 1, \\2a_{n},n = 2k,\end{cases}k \in N^{*}$，则$b_{1} = a_{1} - 6$，$b_{2} = 2a_{2} = 2a_{1} + 2d$，$b_{3} = a_{3} - 6 = a_{1} + 2d - 6$，于是$\begin{cases}S_{4} = 4a_{1} + 6d = 32, \\T_{3} = 4a_{1} + 4d - 12 = 16,\end{cases}$解得$a_{1} = 5$，$d = 2$，$a_{n} = a_{1} + (n - 1)d = 2n + 3$，所以数列$\{ a_{n}\}$的通项公式是$a_{n} = 2n + 3$。
(2)证明：法一：由
(1)知，$S_{n} = \frac{n(5 + 2n + 3)}{2} = n^{2} + 4n$，$b_{n} = \begin{cases}2n - 3,n = 2k - 1, \\4n + 6,n = 2k,\end{cases}k \in N^{*}$，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，$T_{n} - S_{n} = (\frac{3}{2}n^{2} + \frac{7}{2}n) - (n^{2} + 4n) = \frac{1}{2}n(n - 1) ＞ 0$，因此$T_{n} ＞ S_{n}$；当$n$为奇数时，$T_{n} = T_{n + 1} - b_{n + 1} = \frac{3}{2}(n + 1)^{2} + \frac{7}{2}(n + 1) - [4(n + 1) + 6] = \frac{3}{2}n^{2} + \frac{5}{2}n - 5$，当$n ＞ 5$时，$T_{n} - S_{n} = (\frac{3}{2}n^{2} + \frac{5}{2}n - 5) - (n^{2} + 4n) = \frac{1}{2}(n + 2)(n - 5) ＞ 0$，因此$T_{n} ＞ S_{n}$。所以当$n ＞ 5$时，$T_{n} ＞ S_{n}$。
法二：由
(1)知，$S_{n} = \frac{n(5 + 2n + 3)}{2} = n^{2} + 4n$，$b_{n} = \begin{cases}2n - 3,n = 2k - 1, \\4n + 6,n = 2k,\end{cases}k \in N^{*}$，当$n$为偶数时，$T_{n} = (b_{1} + b_{3} + ·s + b_{n - 1}) + (b_{2} + b_{4} + ·s + b_{n}) = -1 + 2(n - 1) - 3 + \frac{n}{2} + \frac{14 + 4n + 6}{2} · \frac{n}{2} = \frac{3}{2}n^{2} + \frac{7}{2}n$，当$n ＞ 5$时，$T_{n} - S_{n} = (\frac{3}{2}n^{2} + \frac{7}{2}n) - (n^{2} + 4n) = \frac{1}{2}n(n - 1) ＞ 0$，因此$T_{n} ＞ S_{n}$；当$n$为奇数时，若$n \geqslant 3$，则$T_{n} = (b_{1} + b_{3} + ·s + b_{n}) + (b_{2} + b_{4} + ·s + b_{n - 1}) = -1 + 2n - 3 + \frac{n + 1}{2} + \frac{14 + 4(n - 1) + 6}{2} · \frac{n - 1}{2} = \frac{3}{2}n^{2} + \frac{5}{2}n - 5$，显然$T_{1} = b_{1} = -1$满足上式，因此当$n$为奇数时，$T_{n} = \frac{3}{2}n^{2} + \frac{5}{2}n - 5$，当$n ＞ 5$时，$T_{n} - S_{n} = (\frac{3}{2}n^{2} + \frac{5}{2}n - 5) - (n^{2} + 4n) = \frac{1}{2}(n + 2)(n - 5) ＞ 0$，因此$T_{n} ＞ S_{n}$。所以当$n ＞ 5$时，$T_{n} ＞ S_{n}$。