《2025年练习生高中数学选择性必修第二册人教版》

1.数列$\frac{3}{5},\frac{4}{7},\frac{5}{9},\frac{6}{11}$，⋯，则该数列的第$n$项为(D)

A.$\frac{n}{2n-1}$
B.$\frac{n+2}{2n-3}$
C.$\frac{n}{2n+1}$
D.$\frac{n+2}{2n+3}$

答案： 1.D 【解析】设该数列为$\{ a_{n}\}$，则$a_{1}=\frac {3}{5},a_{2}=\frac {4}{7}=\frac {3 + 1}{5 + 2}$，$a_{3}=\frac {5}{9}=\frac {3 + 1 × 2}{5 + 2 × 2},a_{4}=\frac {6}{11}=\frac {3 + 1 × 3}{5 + 2 × 3},·s$，以此类推，可得$a_{n}=\frac {3 + (n - 1)}{5 + 2(n - 1)}=\frac {n + 2}{2n + 3}$.故选D.

2.在等差数列$\{ a_n \}$中，若$a_2 + a_6 = 12$，$a_7 = 15$，则$a_{10}$等于
(D)

A.20
B.18
C.16
D.24

答案： 2.D 【解析】因为数列$\{ a_{n}\}$是等差数列，所以$a_{2}+a_{6}=2a_{4}=12$，所以$a_{4}=6$.又$a_{7}=15$，所以$a_{4}+a_{10}=2a_{7}=30$，所以$a_{10}=30 - a_{4}=24$.故选D.

3.已知数列$\{ a_n \}$的前$n$项和为$S_n$.若$a_1 = 2$，$a_{n + 1} = S_n$，则$a_{100} =$(C)

A.$2^{97}$
B.$2^{98}$
C.$2^{99}$
D.$2^{100}$

答案： 3.C 【解析】当$n \geqslant 2$时，由$a_{n + 1}=S_{n}$①，可得$a_{n}=S_{n - 1}$②，两式相减得$a_{n + 1}-a_{n}=a_{n}(n \geqslant 2)$，所以$a_{n + 1}=2a_{n}(n \geqslant 2)$.当$n = 1$时，$a_{2}=S_{1}=a_{1}=2$，所以数列$\{ a_{n}\}$从第二项开始是公比为$2$的等比数列，所以$a_{n}=\begin{cases} 2,n = 1, \\ 2^{n - 1},n \geqslant 2, \end{cases}$所以$a_{100}=2^{99}$.故选C.

4.已知数列$\{ a_n \}$满足$a_{n + 1} = \frac{1}{2}a_n$.若$a_4 + a_5 = 3$，则$a_2 + a_3 =$(D)

A.$\frac{1}{9}$
B.1
C.6
D.12

答案： 4.D 【解析】由$a_{n + 1}=\frac {1}{2}a_{n}$知，数列$\{ a_{n}\}$是公比$q$为$\frac {1}{2}$的等比数列，所以$a_{4}+a_{5}=a_{2}q^{2}+a_{3}q^{2}=(a_{2}+a_{3})q^{2}=\frac {1}{4}(a_{2}+a_{3}) = 3$，解得$a_{2}+a_{3}=12$.故选D.

5.[2022·江苏连云港灌南高级中学高二月考]定义：在数列$\{ a_n \}$中，若满足$\frac{a_{n + 2}}{a_{n + 1}} - \frac{a_{n + 1}}{a_n} = d ( n \in \mathbf{N}^* , d$为常数)，则称$\{ a_n \}$为“等差比数列”.已知在“等差比数列”$\{ a_n \}$中，$a_1 = a_2 = 1$，$a_3 = 3$，则$\frac{a_{2\ 021}}{a_{2\ 019}} =$(C)

A.$4 × 2\ 021^2 - 1$
B.$4 × 2\ 020^2 - 1$
C.$4 × 2\ 019^2 - 1$
D.$4 × 2\ 019^2$

答案： 5.C 【解析】由题意，得$d = \frac {a_{3}}{a_{2}} - \frac {a_{2}}{a_{1}} = 3 - 1 = 2$，所以$\frac {a_{n + 1}}{a_{n}} = \frac {a_{2}}{a_{1}} + 2(n - 1) = 2n - 1$，$\frac {a_{n + 2}}{a_{n + 1}} = 2n + 1$，所以$\frac {a_{n + 2}}{a_{n}} = \frac {a_{n + 2}}{a_{n + 1}} · \frac {a_{n + 1}}{a_{n}} = (2n - 1)(2n + 1)$，即$\frac {a_{n + 2}}{a_{n}} = 4n^{2} - 1$.故选C.

6.风雨桥[如图(1)]是侗族最具特色的民间建筑之一.风雨桥由桥、塔、亭组成.其中亭、塔的俯视图通常是正方形、正六边形或正八边形.图(2)是某风雨桥亭的大致俯视图，其中正六边形的边长的计算方法如下：$A_1 B_1 = A_0 B_0 - B_0 B_1$，$A_2 B_2 = A_1 B_1 - B_1 B_2$，⋯，$A_n B_n = A_{n - 1} B_{n - 1} - B_{n - 1} B_n$，其中$B_3 B_4 = B_2 B_3 = B_1 B_2 = B_0 B_1$，$n \in \mathbf{N}^*$.已知该风雨桥亭共5层，若$A_0 B_0 = 8\ m$，$B_0 B_1 = 0.5\ m$，则图(2)中的五个正六边形的周长总和为(B)

A.$120\ m$
B.$210\ m$
C.$130\ m$
D.$310\ m$

答案： 6.B 【解析】由题意，得$A_{n}B_{n}=A_{n - 1}B_{n - 1}-B_{n - 1}B_{n}(n \leqslant 4$且$n \in \mathbf{N}^{*})$，$B_{1}B_{2}=B_{2}B_{3}=B_{3}B_{4}=B_{4}B_{5}=0.5\mathrm{m}$，故易知题图
(2)中五个正六边形的边长（单位：m）构成以$a_{1}=8$为首项，$d = - 0.5$为公差的等差数列$\{ a_{k}\}$.设数列$\{ a_{k}\}(k \in \mathbf{N}^{*},1 \leqslant k \leqslant 5)$的前$5$项和为$S_{5}$，则$S_{5}=5a_{1}+\frac {1}{2} × 5 × 4 × d = 5 × 8 - \frac {1}{2} × 5 × 4 × 0.5 = 35$，所以题图
(2)中的五个正六边形的周长总和为$6S_{5}=6 × 35 = 210(\mathrm{m})$.故选B.

7.[2022·湖南长沙明德中学高二月考]数列$\{ b_n \}$为正项等比数列，且$b_1 = 1$.等差数列$\{ a_n \}$的首项$a_1 = 2$，且$a_2 = b_3$，$a_4 = b_4$.记$c_n = \frac{a_n}{b_n}$，数列$\{ c_n \}$的前$n$项和为$S_n$，对$\forall n \in \mathbf{N}^*$，$k ＞ S_n \geqslant t ( k , t \in \mathbf{Z} )$恒成立，则$k - t$的最小值为(C)

A.2
B.4
C.6
D.8

答案： 7.C 思维思路 分别设出等差数列的公差和等比数列的公比→解方程可得公差和公比→得$a_{n},b_{n}$→利用错位相减法求和→得$S_{n}$→结合不等式可求得$k - t$的最小值.
【解析】设正项等比数列$\{ b_{n}\}$的公比为$q$，则$q ＞ 0$，等差数列$\{ a_{n}\}$的公差为$d$，则由$a_{2}=b_{3},a_{4}=b_{4}$得$2 + d = q^{2},2 + 3d = q^{3}$，得$q^{3}-3q^{2}+4 = 0$，即$(q + 1)(q - 2)^{2}=0$，解得$q = 2$或$q = - 1$（舍去），则$d = 2$.所以$a_{n}=2 + 2(n - 1)=2n$，$b_{n}=2^{n - 1}$，所以$c_{n}=\frac {a_{n}}{b_{n}}=n · (\frac {1}{2})^{n - 2}$（数列求和，此处用错位相减法较合适），所以$S_{n}=1 · (\frac {1}{2})^{-1}+2 · (\frac {1}{2})^{0}+·s+(n - 1) · (\frac {1}{2})^{n - 3}+n · (\frac {1}{2})^{n - 2}$，$\frac {1}{2}S_{n}=1 · (\frac {1}{2})^{0}+2 · (\frac {1}{2})^{-1}+·s+(n - 1) · (\frac {1}{2})^{n - 2}+n · (\frac {1}{2})^{n - 1}$，两式相减可得$-\frac {1}{2}S_{n}=2+(\frac {1}{2})^{0}+(\frac {1}{2})^{+}·s+(\frac {1}{2})^{n - 3}+(\frac {1}{2})^{n - 2}-n · (\frac {1}{2})^{n - 1}=2+1 - (\frac {1}{2})^{n - 1}1 - \frac {1}{2}-n · (\frac {1}{2})^{n - 1}$，整理得$S_{n}=8 - (n + 2) · (\frac {1}{2})^{n - 2}$.
由$S \geqslant S_{5}=2,S_{6} ＜ 8$，可得$2 \leqslant S_{n} ＜ 8$.因为$S_{n}$对$\forall n \in \mathbf{N}^{*},k ＞ S_{n} \geqslant t(k,t \in \mathbf{Z})$恒成立，所以$k \geqslant 8,t \leqslant 2$，所以$k$的最小值为$8$，$t$的最大值为$2$，$k - t$的最小值为$6$.故选C.

8.已知数列$\{ a_n \}$满足$a_1 = 1$，$a_{n + 1} = \frac{a_n}{1 + \sqrt{a_n}} ( n \in \mathbf{N}^* )$，记$S_n$为数列$\{ a_n \}$的前$n$项和，则(A)

A.$2 ＜ S_{50} ＜ 3$
B.$\frac{3}{2} ＜ S_{50} ＜ 3$
C.$3 ＜ S_{50} ＜ 4$
D.$4 ＜ S_{50} ＜ \frac{9}{2}$

答案： 8.A 【解析】因为$a_{1}=1,a_{n + 1}=\frac {a_{n}}{1 + \sqrt {a_{n}}}$，所以$a_{n} ＞ 0,a_{2}=\frac {1}{2}$，所以$S_{50} ＞ a_{1}+a_{2}=\frac {3}{2},a_{n + 1}=\frac {1}{a_{n}^{-1}+\sqrt {a_{n}}}(\frac {1}{\sqrt {a_{n}}} - \frac {1}{\sqrt {a_{n}+1}})=\frac {1}{\sqrt {a_{n}}}(\frac {1}{\sqrt {a_{n}}} - \frac {1}{\sqrt {a_{n}+1}})$，所以$\frac {1}{\sqrt {a_{n + 1}}} ＜ \frac {1}{\sqrt {a_{n}}}=\frac {1}{2}$，故当$n \geqslant 2$时，$\frac {1}{\sqrt {a_{n}}} ＜ \frac {1}{\sqrt {a_{2}}}+\sum_{i = 2}^{n - 1}(\frac {1}{\sqrt {a_{i + 1}}} - \frac {1}{\sqrt {a_{i}}}) ＜ \frac {1}{\sqrt {a_{2}}}=\sqrt {2}$.当$n \geqslant 2$时，由累加法可得$\frac {1}{\sqrt {a_{n}}} ＜ 1+\frac {n - 1}{2}=\frac {n + 1}{2}$，即$a_{n} ＞ \frac {4}{(n + 1)^{2}}$.当$n = 1$时，$a_{1}=\frac {4}{(1 + 1)^{2}}$成立，所以$a_{n} \geqslant \frac {4}{(n + 1)^{2}}(n \in \mathbf{N}^{*})$，所以$a_{n + 1}=\frac {a_{n}}{1 + \sqrt {a_{n}}} \leqslant \frac {a_{n}}{1 + \frac {2}{n + 1}}=\frac {n + 1}{n + 3}a_{n}$，所以$\frac {a_{n + 1}}{a_{n}} \leqslant \frac {n + 1}{n + 3}$，故当$n \geqslant 2$时，$\frac {a_{n}}{a_{1}} \leqslant \frac {a_{2}}{a_{1}} · \frac {a_{3}}{a_{2}} · \frac {a_{4}}{a_{3}} ·s \frac {a_{n}}{a_{n - 1}} \leqslant \frac {2}{4} × \frac {3}{5} × \frac {4}{6} × ·s × \frac {n}{n + 2}=\frac {1 × 2 × 3 × 3 × 4 × ·s × n}{4 × 5 × 6 × ·s × (n + 2)}=\frac {6}{ (n + 1)(n + 2)}$，所以$S_{50} \leqslant 6 × (\frac {1}{2 × 3}+\frac {1}{3 × 4}+\frac {1}{4 × 5}+·s+\frac {1}{51 × 52}) = 6 × (\frac {1}{2} - \frac {1}{52}) ＜ 3$，所以$S_{50} ＜ 3$.因为$a_{n} \geqslant \frac {4}{(n + 1)^{2}} ＞ \frac {4}{(n + 1)(n + 2)} = 4(\frac {1}{n + 1} - \frac {1}{n + 2})$，所以$S_{50}=a_{1}+a_{2}+·s+a_{50} ＞ 1+\frac {1}{2}+4 × (\frac {1}{3} - \frac {1}{4}+\frac {1}{4} - \frac {1}{5}+·s+\frac {1}{51} - \frac {1}{52}) = \frac {3}{2}+\frac {1}{3} - \frac {1}{13} ＞ 2$，所以$2 ＜ S_{50} ＜ 3$.故选A.

9.设数列$\{ a_n \}$的前$n$项和为$S_n$，下列命题正确的是
(AC)

A.若$\{ a_n \}$为等差数列，则$S_n$，$S_{2n} - S_n$，$S_{3n} - S_{2n}$仍为等差数列
B.若$\{ a_n \}$为等比数列，则$S_n$，$S_{2n} - S_n$，$S_{3n} - S_{2n}$仍为等比数列
C.若$\{ a_n \}$为等差数列，则$\{ a^a_n \}$($a$为常数且$a ＞ 0$)为等比数列
D.若$\{ a_n \}$为等比数列，则$\{ \lg a_n \}$为等差数列

答案： 9.AC 【解析】对于A，若$\{ a_{n}\}$为等差数列，则$S_{n}=a_{1}n+\frac {n(n - 1)d}{2},S_{2n}=2a_{1}n+n(2n - 1)d,S_{3n}=3a_{1}n+\frac {3n(3n - 1)d}{2}$，$S_{n}+S_{3n}-S_{2n}=a_{1}n+\frac {n(n - 1)d}{2}+3a_{1}n+\frac {3n(3n - 1)d}{2}-2a_{1}n-n(2n - 1)d=a_{1}n+\frac {3n^{2}d}{2}-\frac {nd}{2}d$，所以$S_{n},S_{2n}-S_{n},S_{3n}-S_{2n}$仍为等差数列，A正确；对于B，若$\{ a_{n}\}$为等比数列，不妨设$a_{1}=2,q = - 1$，则$S_{2}=a_{1}+a_{2}=0,S_{4}=0,S_{6}=0$，则$S_{2},S_{4}-S_{2},S_{6}-S_{4}$不为等比数列，B错误；对于C，若$\{ a_{n}\}$为等差数列，设$a_{n}=a_{1}+(n - 1)d$，则$a^{a_{n + 1}}=a^{a_{1}+nd}=a^{a_{1}} · (a^{d})^{n},a^{a_{n}}=a^{a_{1}+(n - 1)d}=a^{a_{1}} · (a^{d})^{n - 1}=a^{a^{d}}$，$a^{d}$为不等于$0$的定值，所以$\{ a^{a_{n}}\}(a$为常数且$a ＞ 0)$为等比数列，C正确；对于D，若$\{ a_{n}\}$为等比数列，设$a_{n}=a_{1}q^{n - 1}$，当$a_{1} ＜ 0,q ＞ 0$时，$a_{n}=a_{1}q^{n - 1} ＜ 0$，此时$\lg a_{n}$无意义，D错误.故选AC.