《2025年小题狂做高中数学选择性必修第二册人教版》

1. (教材变式)已知等比数列$\{ a_{n}\}$的前n项和为$S_{n}$,满足$a_{1}=1$,公比$q=2$.若$S_{n}=31$,则$n=$ ( )

A.4
B.5
C.6
D.7

答案： 1. B 由$31 = \frac{1 - 2^{n}}{1 - 2}$得$2^{n} = 32$，解得$n = 5$.
教材链接 人教A版选择性必修二4.3.2例7

2. 设数列$\{ (-1)^{n}\}$的前n项和为$S_{n}$,则$S_{n}=$ ( )

A.$\frac {n[(-1)^{n}-1]}{2}$
B.$\frac {(-1)^{n+1}+1}{2}$
C.$\frac {(-1)^{n}+1}{2}$
D.$\frac {(-1)^{n}-1}{2}$

答案： 2. D $S_{n} = \frac{(-1)[1 - (-1)^{n}]}{1 - (-1)} = \frac{(-1)^{n} - 1}{2}$

3. (易错易混)在等比数列$\{ a_{n}\}$中,$a_{3}=\frac {3}{2},S_{3}=\frac {9}{2}$,则$a_{1}=$ ( )

A.$\frac {3}{2}$或6
B.3
C.$\frac {3}{2}$或3
D.6

答案： 3. A 设等比数列$\{ a_{n}\}$的公比为$q$，由题意可得$a_{3} = a_{1}q^{2} = \frac{3}{2}$，
解得$\begin{cases} a_{1} = 6, \\q = \frac{1}{2} \end{cases}$或$\begin{cases} a_{1} = \frac{3}{2}, \\q = - 1 \end{cases}$或$\begin{cases} a_{1} = 6, \\q = 1 \end{cases}$或$\begin{cases} a_{1} = \frac{3}{2}, \\q = - \frac{1}{2} \end{cases}$.
$S_{3} = a_{1}(1 + q + q^{2}) = \frac{9}{2}$，
易错警示 等比数列求和的技巧
涉及等比数列的前$n$项和，若项数较小，则可直接相加，不需要用等比数列的前$n$项和公式；若用前$n$项和公式，则必须讨论公比$q = 1$和$q \neq 1$.

4. 设等比数列$\{ a_{n}\}$的前n项和为$S_{n}$.若$\frac {S_{6}}{S_{3}}=3$,则$\frac {S_{9}}{S_{6}}=$ ( )

A.2
B.$\frac {7}{3}$
C.$\frac {8}{3}$
D.3

答案： 4. B 设公比为$q$，因为$S_{6} = 3S_{3} \neq 2S_{3}$，所以$q \neq 1$，则$\frac{S_{6}}{S_{3}} = \frac{1 - q^{6}}{1 - q^{3}} = 1 + q^{3} = 3$，得$q^{3} = 2$，于是$\frac{S_{9}}{S_{6}} = \frac{1 - q^{9}}{1 - q^{6}} = \frac{1 - 2^{3}}{1 - 2^{2}} = \frac{7}{3}$.

5. 已知一个等比数列的首项为1,项数是偶数,其奇数项之和为85,偶数项之和为170,则这个数列的项数为 ( )

A.2
B.4
C.8
D.16

答案： 5. C 设这个等比数列$\{ a_{n}\}$共有$2k(k \in \mathbf{N}^{*})$项，公比为$q$，则奇数项之和为$S_{奇} = a_{1} + a_{3} + ·s + a_{2k - 1} = 85$，偶数项之和为$S_{偶} = a_{2} + a_{4} + ·s + a_{2k} = q(a_{1} + a_{3} + ·s + a_{2k - 1}) = qS_{奇} = 170$，所以$q = \frac{S_{偶}}{S_{奇}} = \frac{170}{85} = 2$，则等比数列$\{ a_{n}\}$的所有项之和为$S_{2k} = \frac{a_{1}(1 - 2^{2k})}{1 - 2} = 2^{2k} - 1 = 170 + 85 = 255$，所以$2^{2k} = 256$，解得$k = 4$，因此，项数为8.

6. 在等比数列$\{ a_{n}\}$中,已知$a_{1}+a_{2}+... +a_{10}=48,a_{1}a_{2}· ... · a_{10}=32$,则$\frac {1}{a_{1}}+\frac {1}{a_{2}}+... +\frac {1}{a_{10}}=$ ( )

A.12
B.16
C.24
D.36

答案： 6. C 设等比数列$\{ a_{n}\}$的公比为$q$，则$\{\frac{1}{a_{n}}\}$也是等比数列，公比为$\frac{1}{q}$.因为$a_{1} · a_{2} · ·s · a_{10} = 32$，即$a_{1}^{10}q^{45} = (a_{1}^{2}q^{9})^{5} = 32$，所以$a_{1}^{2}q^{9} = 2$.又因为$a_{1} + a_{2} + ·s + a_{10} = 48$，可设$\frac{1}{a_{1}} + \frac{1}{a_{2}} + ·s + \frac{1}{a_{10}} = x$，则$\frac{1}{a_{1}}\left[1 - \left(\frac{1}{q}\right)^{10}\right]\frac{1}{1 - \frac{1}{q}} = x$，
整理得$\begin{cases} a_{1}(1 - q^{10}) = 48, \\ \frac{1 - q^{10}}{a_{1}q^{9}(q - 1)} = x, \end{cases}$两式相除得$a_{1}^{2}q^{9} = \frac{48}{x}$，则$x = \frac{48}{a_{1}^{2}q^{9}} = \frac{48}{2} = 24$，即$\frac{1}{a_{1}} + \frac{1}{a_{2}} + ·s + \frac{1}{a_{10}} = 24$.

7. 等比数列$\{ a_{n}\}$的公比为$q(q＜0)$,且$a_{4},a_{3},a_{5}$成等差数列,则下列说法正确的是 ( )

A.$q=-2$
B.若$a_{1}=1$,则$a_{2}a_{8}=256$
C.若$S_{2}=-1$,则$S_{4}=-4$
D.$\frac {S_{9}-S_{6}}{S_{6}-S_{3}}=-2$

答案： 7. AB 对于A，因为$a_{4},a_{3},a_{5}$成等差数列，所以$2a_{3} = a_{4} + a_{5}$，即$2a_{3} = a_{3}q + a_{3}q^{2}$.又因为$a_{3} \neq 0$，所以$2 = q + q^{2}$，解得$q = 1$或$q = - 2$，而$q ＜ 0$，所以$q = - 2$，故A正确.对于B，因为$a_{1} = 1$，所以$a_{2}a_{8} = a_{5}^{2} = (a_{1}q^{4})^{2} = 2^{8} = 256$，故B正确.
对于C，因为$S_{2} = - 1$，所以$S_{2} = a_{1} + a_{2} = - 1$，所以$S_{4} = a_{1} + a_{2} + a_{3} + a_{4} = (1 + q^{2})S_{2} = - 5$，故C错误.对于D，
$\frac{S_{9} - S_{6}}{S_{6} - S_{3}} = \frac{a_{7} + a_{8} + a_{9}}{a_{4} + a_{5} + a_{6}} = q^{3} = - 8$，故D错误.

8. 已知等比数列$\{ a_{n}\}$的公比为q,前n项和为$S_{n}$,则下列结论正确的是 ( )

A.$S_{n+1}=S_{n}+a_{n}· q$
B.$S_{n+1}=S_{1}+qS_{n}$
C.$\frac {1}{a_{1}}+\frac {1}{a_{2}}+... +\frac {1}{a_{n}}=\frac {1}{a_{n}}· \frac {1-q^{n}}{1-q}$
D.“$q=-\frac {1}{2}$”是“$S_{n},S_{n+2},S_{n+1}$成等差数列”的充要条件

答案： 8. ABD 对于A，$S_{n + 1} - S_{n} = a_{n + 1} = a_{1}q^{n} = a_{n} · q$，故A正确.对于B，$S_{n + 1} - S = a_{2} + a_{3} + ·s + a_{n + 1} = q(a_{1} + a_{2} + ·s + a_{n}) = qS_{n}$，故B正确.对于C，$\frac{1}{a_{n}} · \frac{1}{a_{n - 1}} · ·s · \frac{1}{a_{2}} · \frac{1}{a_{1}}$可看作是以$\frac{1}{a_{n}}$为首项，$q$为公比的等比数列，当$q \neq 1$时，$\frac{1}{a_{1}} + \frac{1}{a_{2}} + ·s + \frac{1}{a_{n}} = \frac{1}{a_{n}} · \frac{1 - q^{n}}{1 - q}$，当$q = 1$时，则上式不成立，故C错误.对于D，当$q = - \frac{1}{2}$时，$S_{n + 1} + S_{n} = 2S_{n + 2} - 2a_{n + 2} - a_{n + 1} = 2S_{n + 2} - (2q · a_{n + 1} + a_{n + 1}) = 2S_{n + 2} - (- a_{n + 1} + a_{n + 1}) = 2S_{n + 2}$.当$S_{n},S_{n + 2},S_{n + 1}$成等差数列时，$S_{n} + S_{n + 1} = 2S_{n + 2}$，即$S_{n} + S_{n + 1} = 2S_{n + 2} = 2S_{n + 1} + 2a_{n + 2} + a_{n + 1}$，即$2a_{n + 2} + a_{n + 1} = 0$，所以$\frac{a_{n + 2}}{a_{n + 1}} = q = - \frac{1}{2}$.综上，“$q = - \frac{1}{2}$”是“$S_{n},S_{n + 2},S_{n + 1}$成等差数列”的充要条件，故D正确.