《2025年优化探究同步导学案高中数学选择性必修第二册人教版》

[例1] 在等比数列$\{a_{n}\}$中，
(1)$ a_{1}+a_{3}=10 $，$ a_{4}+a_{6}=\dfrac{5}{4} $，求$ S_{5} $；
(2)$ S_{2}=30 $，$ S_{3}=155 $，求$ S_{n} $；
(3)$ a_{1}+a_{n}=66 $，$ a_{2}a_{n - 1}=128 $，$ S_{n}=126 $，求公比$ q $.

答案：
[例1][解]设等比数列$\{ a_{n}\}$的公比为$q$.
(1)法一：由题意，知$\begin{cases}a_{1} + a_{1}q^{2} = 10,\\a_{1}q^{3} + a_{1}q^{5} = \frac{5}{4}.\end{cases}$解得$\begin{cases}a_{1} = 8,\\q = \frac{1}{2}.\end{cases}$从而$S_{5} = \frac{a_{1}(1 - q^{5})}{1 - q} = \frac{31}{2}$.
法二：由$(a_{1} + a_{3})q^{3} = a_{4} + a_{6}$，得$q^{3} = \frac{1}{8}$，从而$q = \frac{1}{2}$.
又因为$a_{1} + a_{3} = a_{1}(1 + q^{2}) = 10$，所以$a_{1} = 8$，从而$S_{5} = \frac{a_{1}(1 - q^{5})}{1 - q} = \frac{31}{2}$.
(2)由题意，知$\begin{cases}a_{1}(1 + q) = 30,\\a_{1}(1 + q + q^{2}) = 155.\end{cases}$
解得$\begin{cases}a_{1} = 5,\\q = 5\end{cases}$或$\begin{cases}a_{1} = 180,\\q = - \frac{5}{6}.\end{cases}$
从而$S_{n} = \frac{1}{4} × 5^{n + 1} - \frac{5}{4}$或$S_{n} = 1080×\left[1 - \left(- \frac{5}{6}\right)^{n}\right]×\frac{11}{1}$
(3)因为$a_{2}a_{n - 1} = a_{1}a_{n} = 128$，所以$a_{1},a_{n}$是方程$x^{2} - 66x + 128 = 0$的两个根，
从而$\begin{cases}a_{1} = 2,\\a_{n} = 64.\end{cases}$或$\begin{cases}a_{1} = 64,\\a_{n} = 2.\end{cases}$
又因为$S_{n} = \frac{a_{1} - a_{n}q}{1 - q} = 126$，所以$q = 2$或$\frac{1}{2}$.

1. (多选)已知在等比数列$\{a_{n}\}$中，$ a_{3}=7 $，前三项之和$ S_{3}=21 $，则$ S_{5} $的值可能是(AC)

A.$ 35 $
B.$ -\dfrac{1}{2} $
C.$ \dfrac{77}{4} $
D.$ 1 $

答案： 1.AC　设等比数列$a_{n} = a_{1}q^{n - 1}(q\neq 0)$，则有$a_{3} = a_{1}q^{2} = 7$，即$a_{1} = \frac{7}{q^{2}}$，$S_{3} = a_{1} + a_{2} + a_{3} = a_{1}(1 + q + q^{2}) = \frac{7(1 + q + q^{2})}{q^{2}}$，即$2q^{2} - q - 1 = 0$，解得$q = 1$或$q = - \frac{1}{2}$，
当$q = 1$时，则$a_{n} = a_{3} = 7$，故$S_{5} = 5×7 = 35$；
当$q = - \frac{1}{2}$时，$a_{n} = a_{3}q^{n - 3} = 7×\left(- \frac{1}{2}\right)^{n - 3} = 28×\left(- \frac{1}{2}\right)^{n - 1}$，则$a_{1} = 28$，$S_{5} = \frac{a_{1}(1 - q^{5})}{1 - q} = \frac{28×\left[1 - \left(- \frac{1}{2}\right)^{5}\right]}{1 - \left(- \frac{1}{2}\right)} = \frac{77}{4}×\frac{3}{2}$

2. 已知等比数列$\{a_{n}\}$的各项均为正数，其前$ n $项和为$ S_{n} $，若$ a_{2}a_{4}=1 $，$ S_{3}=7 $，则$ S_{5}=$$\frac{31}{4}$.

答案： 2.答案：$\frac{31}{4}$
解析：由题知等比数列$\{ a_{n}\}$的各项均为正数，即$a_{n}＞0$，设其公比为$q$，则$q＞0$，
因为$a_{2}a_{4} = 1$，$S_{3} = 7$，所以$a_{1}q· a_{1}q^{3} = a_{1}^{2}q^{4} = 1$，即$a_{1} = \frac{1}{q^{2}}$，$S_{3} = a_{1} + a_{2} + a_{3}$，即$a_{1}(1 + q + q^{2}) = 7$，即$\frac{1}{q^{2}}(1 + q + q^{2}) = 7$，即$6q^{2} - q - 1 = 0$，即$(3q + 1)(2q - 1) = 0$，
故$q = \frac{1}{2}$或$q = - \frac{1}{3}$(舍去)，则$a_{1} = \frac{1}{q^{2}} = 4$，所以$S_{5} = \frac{4×\left[1 - \left(\frac{1}{2}\right)^{5}\right]}{1 - \frac{1}{2}} = \frac{31}{4}$.