《2025年热搜题高中数学选择性必修第二册人教版》

1. [2024·邯郸一中月考]已知数列$\{ a_{n}\} $是公差不为零的等差数列，且$a_{5},a_{8},a_{13}$是等比数列$\{ b_{n}\} $中的连续三项，若$b_{2}=2$，则$b_{n}=$(A)

A.$2×\left( \dfrac{5}{3}\right)^{n-2}$
B.$2×\left( \dfrac{2}{3}\right)^{n-2}$
C.$2×\left( \dfrac{3}{5}\right)^{n-1}$
D.$2×\left( \dfrac{5}{3}\right)^{n-1}$

答案： 1.A 【解析】因为$\{ a_{n}\}$是公差不为零的等差数列，并且$a_{5}$，$a_{8}$，$a_{13}$是等比数列$\{ b_{n}\}$的连续三项，设数列$\{ a_{n}\}$的公差为$d$，则$(a_{5} + 3d)^{2} = a_{5}(a_{5} + 8d)$，化简得$(2a_{5} - 9d)d = 0$。因为$d\neq0$，所以$a_{5} = \frac{9}{2}d$，所以$q = \frac{a_{8}}{a_{5}} = \frac{a_{5} + 3d}{a_{5}} = \frac{\frac{15}{2}d}{\frac{9}{2}d} = \frac{5}{3}$，所以$b_{n} = b_{2}q^{n - 2} = 2 × (\frac{5}{3})^{n - 2}$．故选A．

2. [2024·安阳一中期末]已知$\{ a_{n}\} $为等差数列，其公差为$-2$，且$a_{7}$是$a_{3}$与$a_{9}$的等比中项，$S_{n}$为$\{ a_{n}\} $的前$n$项和，则$S_{10}=$(D)

A.$-110$
B.$-90$
C.$90$
D.$110$

答案： 2.D 【解析】因为$a_{3} = a_{1} + 2d = a_{1} - 4$，$a_{7} = a_{1} + 6d = a_{1} - 12$，$a_{9} = a_{1} + 8d = a_{1} - 16$，且$a_{7}$是$a_{3}$与$a_{9}$的等比中项，所以$(a_{1} - 12)^{2} = (a_{1} - 4)(a_{1} - 16)$，解得$a_{1} = 20$，所以$S_{10} = 10 × 20 + \frac{1}{2} × 10 × 9 × ( - 2) = 110$．故选D.

3. [2024·唐山一中月考]已知等比数列$\{ a_{n}\} $的前$n$项和为$S_{n}$，满足$S_{1},2S_{2},3S_{3}$构成等差数列，且$a_{1}a_{2}=a_{3}$，若$\{ (\log_{3}a_{n})^{2}-\lambda \log_{3}a_{n}\} $是递增数列，则实数$\lambda $的取值范围是$(- 3, + \infty)$.

答案： 3.$(- 3, + \infty)$ 【解析】由题意，得$4S_{2} = S_{1} + 3S_{3}$，化简得$\frac{1}{3}a_{2} = a_{3}$，所以公比$q = \frac{1}{3}$．因为$a_{1}a_{2} = a_{3}$，即$a_{1}^{2}q = a_{1}q^{2}$，而$a_{1} \neq 0$，所以$a_{1} = \frac{1}{3}$，所以$a_{n} = (\frac{1}{3})^{n}$，所以$(\log_{3}a_{n})^{2} - \lambda\log_{3}a_{n} = n^{2} + \lambda n$．因为$\{ n^{2} + \lambda n\}$是递增数列，所以$(n^{2} + \lambda n) - \lbrack(n - 1)^{2} + \lambda(n - 1)\rbrack = 2n + \lambda - 1 ＞ 0$，$n \geqslant 2$，所以$\lambda ＞ - 2n + 1$，得$\lambda ＞ ( - 2n + 1)_{\max} = - 3$．

4. [2024·哈尔滨三中月考]已知单调递增的等比数列$\{ a_{n}\} $的前$n$项和为$S_{n}$，若$S_{3}=39$，且$3a_{4}$是$a_{6}$与$-a_{5}$的等差中项.
（1）求数列$\{ a_{n}\} $的通项公式；
（2）若数列$\{ b_{n}\} $满足$b_{n}=\log_{3}a_{2n+1}$，且$\{ b_{n}\} $的前$n$项和为$T_{n}$，求$\dfrac{1}{T_{1}}+\dfrac{1}{T_{2}}+\dfrac{1}{T_{3}}+·s +\dfrac{1}{T_{n}}$.

答案： 4.
(1)设数列$\{ a_{n}\}$的公比为$q$．由题意，得$6a_{4} = a_{6} - a_{5}$，即$q^{2} - q - 6 = 0$，解得$q = 3$或$q = - 2$．因为该等比数列是单调递增的，所以$q ＞ 1$，故$q = - 2$舍去．
由$S_{3} = \frac{a_{1}(1 - q^{3})}{1 - q} = 39$，得$a_{1} = 3$，所以$a_{n} = 3^{n}$．
(2)$b_{n} = \log_{3}3^{2n + 1} = 2n + 1$，
所以$T_{n} = 3 + 5 + ·s + (2n + 1) = n(n + 2)$，
所以$\frac{1}{T_{n}} = \frac{1}{n(n + 2)} = \frac{1}{2}(\frac{1}{n} - \frac{1}{n + 2})$，
$\frac{1}{T_{1}} + \frac{1}{T_{2}} + \frac{1}{T_{3}} + ·s + \frac{1}{T_{n}} = \frac{1}{2}(1 - \frac{1}{3}) + \frac{1}{2}(\frac{1}{2} - \frac{1}{4}) +$
$\frac{1}{2}(\frac{1}{3} - \frac{1}{5}) + ·s + \frac{1}{2}(\frac{1}{n} - \frac{1}{n + 2}) = \frac{1}{2}(1 + \frac{1}{2} - \frac{1}{n + 1} -$
$\frac{1}{n + 2}) = \frac{1}{2}(\frac{3}{2} - \frac{1}{n + 1} - \frac{1}{n + 2}) = \frac{3}{4} - \frac{1}{2n + 2} - \frac{1}{2n + 4}$．

5. [2024·黄冈中学期末]已知数列$\{ a_{n}\} $的奇数项是首项为$1$，公差为$d$的等差数列，偶数项是首项为$2$，公比为$q$的等比数列，数列$\{ a_{n}\} $的前$n$项和为$S_{n}$，且满足$S_{3}=a_{4}$，$a_{3}+a_{5}=2+a_{4}$.
（1）求$d$和$q$的值；
（2）求$S_{n}$.

答案： 5.
(1)根据题意，得$\begin{cases} S_{3} = a_{1} + a_{2} + a_{3} = a_{4}, \\a_{3} + a_{5} = 2 + a_{4}, \end{cases}$
即$\begin{cases} 1 + 2 + 1 + d = 2q, \\1 + d + 1 + 2d = 2 + 2q, \end{cases}$解得$\begin{cases} d = 2, \\q = 3. \end{cases}$
(2)由
(1)，得当$n$为奇数时，令$n = 2k - 1$，$k \in \mathbf{N}^{*}$，
$a_{2k - 1} = a_{1} + (k - 1)d = 1 + (k - 1) × 2 = 2k - 1$，所以$a_{n} = n$．
当$n$为偶数时，令$n = 2k$，$k \in \mathbf{N}^{*}$，
$a_{2k} = a_{2} · q^{k - 1} = 2 × 3^{k - 1}$，所以$a_{n} = 2 × 3^{\frac{n}{2} - 1}$．
综上所述，$a_{n} = \begin{cases} n,n为奇数, \\2 × 3^{\frac{n}{2} - 1},n为偶数. \end{cases}$
①当$n$为偶数时，数列$\{ a_{n}\}$有$\frac{n}{2}$个奇数项，$\frac{n}{2}$个偶数项，
所以$S_{n} = (a_{1} + a_{3} + ·s + a_{n - 1}) + (a_{2} + a_{4} + ·s + a_{n})$
$= (1 + 3 + ·s + n - 1) + 2 × (1 + 3 + ·s + 3^{\frac{n}{2} - 1})$
$= \frac{1 + n - 1}{2} × \frac{n}{2} + 2 × \frac{1 × (1 - 3^{\frac{n}{2}})}{1 - 3}$
$= \frac{n^{2}}{4} + 3^{\frac{n}{2}} - 1$；
②当$n$为奇数时，$n + 1$为偶数，
所以$S_{n} = S_{n + 1} - a_{n + 1}$
$= \frac{(n + 1)^{2}}{4} + 3^{\frac{n + 1}{2}} - 1 - 2 × 3^{\frac{n - 1}{2}}$
$= \frac{(n + 1)^{2}}{4} + 3^{\frac{n - 1}{2}} - 1$．
综上所述，$S_{n} = \begin{cases} \frac{n^{2}}{4} + 3^{\frac{n}{2}} - 1,n为偶数, \frac{(n + 1)^{2}}{4} + 3^{\frac{n - 1}{2}} - 1,n为奇数. \end{cases}$

6. [2024·河北正定中学月考]已知$\{ a_{n}\} $是一个公差大于$0$的等差数列，且满足$a_{3}a_{6}=55$，$a_{2}+a_{7}=16$.
（1）求数列$\{ a_{n}\} $的通项公式；
（2）若数列$\{ a_{n}\} $和数列$\{ b_{n}\} $满足等式：$a_{n}=\dfrac{b_{1}}{2}+\dfrac{b_{2}}{2^{2}}+\dfrac{b_{3}}{2^{3}}+·s +\dfrac{b_{n}}{2^{n}}$（$n$为正整数），求数列$\{ b_{n}\} $的前$n$项和$S_{n}$.

答案： 6.
(1)设等差数列$\{ a_{n}\}$的公差为$d$，则依题知$d ＞ 0$．
由$a_{2} + a_{7} = 16$，得$2a_{1} + 7d = 16$，
由$a_{3}a_{6} = 55$，得$(a_{1} + 2d)(a_{1} + 5d) = 55$，
由①得$a_{1} = 8 - \frac{7}{2}d$，将其代入②得$(16 - 3d)(16 + 3d) = 220$．
即$256 - 9d^{2} = 220$，整理得$d^{2} = 4$．又$d ＞ 0$，所以$d = 2$．
代入①得$a_{1} = 1$，所以$a_{n} = 1 + (n - 1) × 2 = 2n - 1$，所以$a_{n} = 2n - 1$．
(2)令$c_{n} = \frac{b_{n}}{2^{n}}$，则$a_{n} = c_{1} + c_{2} + ·s + c_{n}$，且$a_{n + 1} = c_{1} +$
$c_{2} + ·s + c_{n + 1}$，
两式相减得$a_{n + 1} - a_{n} = c_{n + 1}$，
由
(1)得$a_{1} = 1$，$d = 2$，$a_{n + 1} - a_{n} = 2$，
则$c_{n + 1} = 2$，即$c_{n} = 2(n \geqslant 2)$，
即当$n \geqslant 2$时，$b_{n} = 2^{n + 1}$．
又当$n = 1$时，$b_{1} = 2a_{1} = 2$，所以$b_{n} = \begin{cases} 2,n = 1, \\2^{n + 1},n \geqslant 2. \end{cases}$
当$n = 1$时，$S_{1} = b_{1} = 2$；
当$n \geqslant 2$时，$S_{n} = b_{1} + b_{2} + b_{3} + ·s + b_{n} = 2 + 2^{3} + 2^{4} + ·s +$
$2^{n + 1} = 2 + 2^{2} + 2^{3} + 2^{4} + ·s + 2^{n + 1} - 4 =$
$\frac{2(1 - 2^{n + 1})}{1 - 2} - 4 =$
$2^{n + 2} - 6$，即$S_{n} = 2^{n + 2} - 6$．
又当$n = 1$时，$2^{3} - 6 = 2$，而$S_{1} = 2$，符合该式，所以$S_{n} =$ $2^{n + 2} - 6$．