《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》

1. 在等比数列$\{a_{n}\}$中，每隔$k$项($k \in \mathbf{N}^{*}$)取出一项，按原来的顺序排列，所得的新数列仍为等比数列.
2. 若$\{a_{n}\}$是等比数列，公比为$q$，则数列$\{\lambda a_{n}\}(\lambda \neq 0)$，$\{\frac{1}{a_{n}}\}$，$\{a_{n}^{2}\}$都是等比数列，且公比分别是$q$，$\frac{1}{q}$，$q^{2}$.
3. 若$\{a_{n}\}$，$\{b_{n}\}$是项数相同的等比数列，公比分别是$p$和$q$，那么$\{a_{n}b_{n}\}$与$\{\frac{a_{n}}{b_{n}}\}$也都是等比数列，公比分别为$pq$和$\frac{p}{q}$.

答案： 3.$pq$

[例4]
已知数列$\{a_{n}\}$的首项$a_{1} = 3$.
(1) 若$\{a_{n}\}$为等差数列，公差$d = 2$，证明数列$\{3^{a_{n}}\}$为等比数列；
(2) 若$\{a_{n}\}$为等比数列，公比$q =\frac{1}{9}$，证明数列$\{\log_{3}a_{n}\}$为等差数列.

答案： [证明]
(1)由$a_{1} = 3,d = 2$，得$\{ a_{n}\}$的通项公式为$a_{n} = 2n + 1$.
设$b_{n} = 3^{a_{n}}$，则$\frac{b_{n + 1}}{b_{n}} = \frac{3^{a_{n + 1}}}{3^{a_{n}}} = \frac{3^{2(n + 1) + 1}}{3^{2n + 1}} = 9$.又因为
$b_{1} = 3^{3} = 27$，
所以，$\{ 3^{a_{n}}\}$是以27为首项，9为公比的等比
数列.
(2)由$a_{1} = 3,q = \frac{1}{9}$，得$a_{n} = 3 × (\frac{1}{9})^{n - 1} = 3^{3 - 2n}$.
两边取以3为底的对数，得$\log_{3}a_{n} = \log_{3}3^{3 - 2n} = 3 - 2n$.
所以$\log_{3}a_{n + 1} - \log_{3}a_{n} = \lbrack 3 - 2(n + 1)\rbrack - (3 - 2n) = - 2$.又因为$\log_{3}a_{1} = \log_{3}3 = 1$，
所以$\{\log_{3}a_{n}\}$是首项为1，公差为$- 2$的等
差数列.

跟踪训练
3. (多选)已知数列$\{a_{n}\}$是等比数列，那么下列数列一定是等比数列的是(AB)

A.$\{\frac{1}{a_{n}}\}$
B.$\{a_{n}a_{n + 1}\}$
C.$\{\lg(a_{n}^{2})\}$
D.$\{a_{n} + a_{n + 1}\}$

答案： 由题意知$\{ a_{n}\}$为等比数
列，设其公比为$q(q \neq 0)$；
对于A，$\frac{1}{a_{n}} = \frac{1}{a_{1}q^{n - 1}} = \frac{1}{a_{1}} · (\frac{1}{q})^{n - 1}$，
$\therefore$数列$\{\frac{1}{a_{n}}\}$是以$\frac{1}{a_{1}}$为首项，$\frac{1}{q}$为公比的等比数列，故A正确；
对于B，$\frac{a_{n + 1}a_{n + 2}}{a_{n}a_{n + 1}} = \frac{a_{n + 2}}{a_{n}} = q^{2}$，$\therefore$数列$\{ a_{n}a_{n + 1}\}$是以$a_{1}a_{2}$为首项，$q^{2}$为公比的等比数列，故B正确；
对于C，当$a_{n} = 1$时，$\lg(a_{n}^{2}) = 0$，数列$\{\lg(a_{n}^{2})\}$不是等比数列，故C错误；
对于D，当$q = - 1$时，$a_{n} + a_{n + 1} = 0$，数列$\{ a_{n} + a_{n + 1}\}$不是等比数列，故D错误.

4. 设$\{a_{n}\}$是各项为正数的无穷数列，$A_{i}$是边长为$a_{i}$，$a_{i + 1}$的矩形面积$(i = 1,2,·s)$，则$\{A_{n}\}$为等比数列的充要条件为(D)

A.$\{a_{n}\}$是等比数列
B.$a_{1},a_{3},·s,a_{2n - 1},·s$或$a_{2},a_{4},·s,a_{2n},·s$是等比数列
C.$a_{1},a_{3},·s,a_{2n - 1},·s$和$a_{2},a_{4},·s,a_{2n},·s$均是等比数列
D.$a_{1},a_{3},·s,a_{2n - 1},·s$和$a_{2},a_{4},·s,a_{2n},·s$均是等比数列，且公比相同

答案： $\because A_{n} = a_{a_{i + 1}}$，若$\{ A_{n}\}$为等比数列，则$\frac{A_{n + 1}}{A_{n}} = \frac{a_{n + 1}a_{n + 2}}{a_{n}a_{n + 1}} = \frac{a_{n + 2}}{a_{n}}$为常数，即$\frac{A_{2}}{A_{1}} = \frac{a_{3}}{a_{1}},\frac{A_{3}}{A_{2} } = \frac{a_{4}}{a_{2}},·s,\therefore a_{1},a_{3},a_{5},·s,a_{2n - 1}·s$和$a_{2},a_{4},·s,a_{2n},·s$均是等比数列，且公比
相等.反之，若奇数项和偶数项分别成等比
数列，且公比相等，设为$q$，则$\frac{A_{n + 1}}{A_{n}} = \frac{a_{n + 2}}{a_{n}} = q$，从而$\{ A_{n}\}$为等比数列.故D正确.