《2025年创新设计高考总复习高三数学人教版A版》

(2025·郑州模拟)已知数列$\{a_{n}\}$的各项均为正数，记$S_{n}$为$\{a_{n}\}$的前n项和，从下面①②③三个条件中选取两个作为已知条件，证明另外一个成立.①数列$\{a_{n}\}$是等比数列；②数列$\{S_{n} + a_{1}\}$是等比数列；③$a_{2} = 2a_{1}$.
注：若选择不同的组合分别解答，按第一个解答计分.证明 选①②作条件证明③.因为数列$\{a_{n}\},\{S_{n}+a_{1}\}$是等比数列,所以$(S_{2}+a_{1})^{2}=(S_{1}+a_{1})(S_{3}+a_{1})$,即$(a_{1}+a_{2}+a_{1})^{2}=(a_{1}+a_{1})(a_{1}+a_{2}+a_{3}+a_{1})$,故$(2a_{1}+a_{2})^{2}=2a_{1}(2a_{1}+a_{2}+a_{3})$,展开得$4a_{1}^{2}+4a_{1}a_{2}+a_{2}^{2}=4a_{1}^{2}+2a_{1}a_{2}+2a_{1}a_{3}$,所以$a_{2}^{2}=2a_{1}a_{3}-2a_{1}a_{2}$,又因为$\{a_{n}\}$是等比数列,设公比为$q$,则$a_{3}=a_{1}q^{2},a_{2}=a_{1}q$,代入得$(a_{1}q)^{2}=2a_{1}(a_{1}q^{2})-2a_{1}(a_{1}q)$,化简得$q^{2}=2q$,解得$q=2$,所以$a_{2}=2a_{1}$.选①③作条件证明②.

答案： 证明 选①②作条件证明③. 因为数列$\{a_{n}\},\{S_{n}+a_{1}\}$是等比数列, 所以$(S_{2}+a_{1})^{2}=(S_{1}+a_{1})(S_{3}+a_{1})$, 即$(a_{1}+a_{2}+a_{1})^{2}=(a_{1}+a_{1})(a_{1}+a_{2}+a_{3}+a_{1})$, 故$(2a_{1}+a_{2})^{2}=2a_{1}(2a_{1}+a_{2}+a_{3})$, 展开得$4a_{1}^{2}+4a_{1}a_{2}+a_{2}^{2}=4a_{1}^{2}+2a_{1}a_{2}+2a_{1}a_{3}$, 所以$a_{2}^{2}=2a_{1}a_{3}-2a_{1}a_{2}$, 又因为$\{a_{n}\}$是等比数列,设公比为$q$,则$a_{3}=a_{1}q^{2},a_{2}=a_{1}q$,代入得$(a_{1}q)^{2}=2a_{1}(a_{1}q^{2})-2a_{1}(a_{1}q)$,化简得$q^{2}=2q$,解得$q=2$,所以$a_{2}=2a_{1}$. 选①③作条件证明②. 因为$a_{2}=2a_{1},\{a_{n}\}$是等比数列, 所以数列$\{a_{n}\}$的公比$q=2$, 所以$S_{n}=\frac{a_{1}(1-2^{n})}{1-2}=a_{1}(2^{n}-1)$, 即$S_{n}+a_{1}=a_{1}\cdot2^{n}$, 因为$\frac{S_{n+1}+a_{1}}{S_{n}+a_{1}}=\frac{a_{1}\cdot2^{n+1}}{a_{1}\cdot2^{n}}=2,S_{n}+a_{1}\neq0$, 所以$\{S_{n}+a_{1}\}$是等比数列. 选②③作条件证明①. 因为数列$\{S_{n}+a_{1}\}$是等比数列,且$a_{2}=2a_{1}$, 所以$\frac{S_{2}+a_{1}}{S_{1}+a_{1}}=\frac{a_{1}+a_{2}+a_{1}}{a_{1}+a_{1}}=\frac{4a_{1}}{2a_{1}}=2$, 则数列$\{S_{n}+a_{1}\}$是以$2a_{1}$为首项,$2$为公比的等比数列, 所以$S_{n}+a_{1}=2a_{1}\cdot2^{n-1}=a_{1}\cdot2^{n}$, 所以$S_{n}=a_{1}\cdot2^{n}-a_{1}$, 当$n\geq2$时,$a_{n}=S_{n}-S_{n-1}=a_{1}\cdot2^{n}-a_{1}-(a_{1}\cdot2^{n-1}-a_{1})=a_{1}\cdot2^{n-1}$, 当$n=1$时,$a_{1}=a_{1}\cdot2^{0}=a_{1}$,符合上式, 所以数列$\{a_{n}\}$是以$a_{1}$为首项,$2$为公比的等比数列.

考点三 等比数列的性质及应用
角度1 项的性质
例3 (2025·重庆七校诊断)已知$\{a_{n}\}$是等差数列，$\{b_{n}\}$是等比数列，若$a_{2} + a_{4} + a_{6} = 4\pi$，$b_{2}b_{4}b_{6} = 3\sqrt{3}$，则$\tan\frac{a_{1} + a_{7}}{1 + b_{2}b_{6}} =$(A)

A.$-\sqrt{3}$
B.$-\frac{\sqrt{3}}{3}$
C.$\frac{\sqrt{3}}{3}$
D.$\sqrt{3}$

答案： A [因为$\{a_{n}\}$是等差数列, 所以$a_{2}+a_{4}+a_{6}=3a_{4}=4\pi$,故$a_{4}=\frac{4\pi}{3}$, 则$a_{1}+a_{7}=2a_{4}=\frac{8\pi}{3}$. 因为$\{b_{n}\}$是等比数列,所以$b_{2}b_{4}b_{6}=b_{4}^{3}=3\sqrt{3}=(\sqrt{3})^{3}$,故$b_{4}=\sqrt{3}$,则$b_{2}b_{6}=b_{4}^{2}=3$, 所以$\tan\frac{a_{1}+a_{7}}{1+b_{2}b_{6}}=\tan\frac{\frac{8\pi}{3}}{4}=\tan\frac{2\pi}{3}=-\sqrt{3}$.]

角度2 和的性质
例4 (2025·大连调研)已知等比数列$\{a_{n}\}$的前n项和为$S_{n}$，若$\frac{S_{4}}{S_{8}} =\frac{1}{4}$，则$\frac{S_{16}}{S_{4} + S_{8}} =$(A)

A.8
B.9
C.16
D.17解 设$S_{4}=x(x\neq0)$,则$S_{8}=4x$.因为$\{a_{n}\}$是等比数列,所以$S_{4},S_{8}-S_{4},S_{12}-S_{8},S_{16}-S_{12}$成等比数列,公比为$\frac{S_{8}-S_{4}}{S_{4}}=3$,则$S_{16}=40x$,故$\frac{S_{16}}{S_{4}+S_{8}}=8$.

答案： A [设$S_{4}=x(x\neq0)$,则$S_{8}=4x$. 因为$\{a_{n}\}$为等比数列,所以$S_{4},S_{8}-S_{4},S_{12}-S_{8},S_{16}-S_{12}$仍成等比数列. 公比为$\frac{S_{8}-S_{4}}{S_{4}}=3$, 所以$S_{12}-S_{8}=9x$,$S_{16}-S_{12}=27x$, 则$S_{16}=40x$, 故$\frac{S_{16}}{S_{4}+S_{8}}=\frac{40x}{x+4x}=8$.]

角度3 等比数列的最值
例5 (多选)设等比数列$\{a_{n}\}$的公比为q，其前n项和为$S_{n}$，前n项积为$T_{n}$，并满足条件$a_{1}$ ＞ 1，$a_{2025}a_{2026}$ ＞ 1，$\frac{a_{2025} - 1}{a_{2026} - 1} ＜ 0$，下列结论正确的是(AB)

A.$S_{2025} ＜ S_{2026}$
B.$a_{2025}a_{2027} - 1 ＜ 0$
C.$T_{2026}$是数列$\{T_{n}\}$中的最大值
D.数列$\{T_{n}\}$无最大值

答案： AB [由$a_{1}＞1$,$a_{2025}a_{2026}＞1$,$\frac{a_{2025}-1}{a_{2026}-1}＜0$知$a_{2025}＞1$,$0＜a_{2026}＜1$,公比$0＜q＜1$. A中,$S_{2026}=S_{2025}+a_{2026}＞S_{2025}$,A正确; B中,$a_{2025}a_{2027}=a_{2026}^{2}＜1$,B正确; C中,$T_{2025}$是最大值,C错误; D中,数列$\{T_{n}\}$有最大值,D错误.]

(1)(2025·蚌埠质检)已知各项均为正数的等比数列$\{a_{n}\}$中，若$a_{5} = 9$，则$\log_{3}a_{4} + \log_{3}a_{6} =$(C)

A.2
B.3
C.4
D.9

答案：
(1)C [
(1)因为$\{a_{n}\}$是等比数列,所以$a_{4}a_{6}=a_{5}^{2}=81$,所以$\log_{3}a_{4}+\log_{3}a_{6}=\log_{3}(a_{4}a_{6})=\log_{3}81=4$.]

(2)(多选)(2025·南京、盐城调研)已知$\{a_{n}\}$是等比数列，$S_{n}$是其前n项和，满足$a_{3} = 2a_{1} + a_{2}$，则下列说法中正确的有(AC)

A.若$\{a_{n}\}$是正项数列，则$\{a_{n}\}$是递增数列
B.$S_{n}$，$S_{2n} - S_{n}$，$S_{3n} - S_{2n}$一定是等比数列
C.若存在M ＞ 0，使$|a_{n}| \leq M$对$n\in N^{*}$都成立，则$\{|a_{n}|\}$是等差数列
D.若存在M ＞ 0，使$|a_{n}| \leq M$对$n\in N^{*}$都成立，则$\{S_{n}\}$是等差数列

答案：
(2)AC [
(2)由$a_{3}=2a_{1}+a_{2}$得$q^{2}-q-2=0$,解得$q=2$或$q=-1$. A中,正项数列则$q=2$,递增数列,A正确; B中,当$q=-1$且$n$为偶数时不成等比数列,B错误; C中,存在$M＞0$则$q=-1$,$|a_{n}|$为常数列即等差数列,C正确; D中,$q=-1$时$S_{n}$不是等差数列,D错误.]

(3)已知数列$\{a_{n}\}$是等比数列，若$a_{2} = 1$，$a_{5} =\frac{1}{8}$，则$a_{1}a_{2} + a_{2}a_{3} + \cdots + a_{n}a_{n + 1}(n\in N^{*})$的最小值为2.

答案：
(3)2 [
(3)公比$q=\frac{1}{2}$,$a_{1}=2$,$a_{n}a_{n+1}=2×\left(\frac{1}{4}\right)^{n-1}$,前$n$项和为$\frac{8}{3}\left[1-\left(\frac{1}{4}\right)^{n}\right]$,最小值为$2$.]