《2024年绿色通道45分钟课时作业与单元测评数学选择性必修第二册》

10.（2023·厦门六中检测）已知数列$\{a_{n}\}$满足：$a_{1}=2,a_{n + 1}=3a_{n}-2,n\in N^{*}$.
(1)设$b_{n}=a_{n}-1$，求数列$\{b_{n}\}$的通项公式；
(2)设$T_{n}=\log_{3}a_{1}+\log_{3}a_{2}+\cdots+\log_{3}a_{n}(n\in N^{*})$，求证：$T_{n}＞\frac{n(n - 1)}{2}$.

答案： 10.解　
(1)因为$a_{n + 1} = 3a_n - 2$，所以$a_{n + 1} - 1 = 3(a_n - 1)$. 又$b_n = a_n - 1$，$a_1 = 2$，所以$b_1 = a_1 - 1 = 1$，$b_{n + 1} = 3b_n$，所以$\frac{b_{n + 1}}{b_n} = 3$，所以$\{b_n\}$是以$1$为首项，$3$为公比的等比数列，所以$b_n = 1\times3^{n - 1} = 3^{n - 1}$.
(2)证明：由
(1)可得$a_n - 1 = 3^{n - 1}$，所以$a_n = 3^{n - 1} + 1$，所以$\log_3a_n = \log_3(3^{n - 1} + 1) \gt \log_33^{n - 1} = n - 1$，所以$T_n = \log_3a_1 + \log_3a_2 + \cdots + \log_3a_n \gt 0 + 1 + 2 + \cdots + (n - 1) = \frac{n(n - 1)}{2}$.

11. 定义：在数列$\{a_{n}\}$中，若满足$\frac{a_{n + 2}}{a_{n + 1}}-\frac{a_{n + 1}}{a_{n}} = d(n\in N^{*},d$为常数$)$，则称$\{a_{n}\}$为“比等差数列”. 已知在“比等差数列”$\{a_{n}\}$中，$a_{1}=a_{2}=1,a_{3}=3$，则$\frac{a_{2023}}{a_{2021}}=$（ ）
A. $4\times2019^{2}-1$
B. $4\times2020^{2}-1$
C. $4\times2021^{2}-1$
D. $4\times2022^{2}-1$

答案： 11.C　由题意得，$\frac{a_2}{a_1} = 1$，$\frac{a_3}{a_2} = 3$，$\therefore\frac{a_3}{a_2} - \frac{a_2}{a_1} = 2$，$\therefore$数列$\{\frac{a_{n + 1}}{a_n}\}$是首项为$1$，公差为$2$的等差数列，$\therefore\frac{a_{n + 1}}{a_n} = 2n - 1$，$\therefore\frac{a_{2023}}{a_{2021}} = \frac{a_{2023}}{a_{2022}}\times\frac{a_{2022}}{a_{2021}} = (2\times2022 - 1)\times(2\times2021 - 1) = (2\times2021 + 1)\times(2\times2021 - 1) = 4\times2021^2 - 1$.

12.（多选）若数列$\{a_{n}\}$的前$n$项和为$S_{n}$，且$S_{n}=2a_{n}+1,n\in N^{*}$，则下列说法正确的是（ ）
A. $a_{5}=-16$
B. $S_{5}=-63$
C. 数列$\{a_{n}\}$是等比数列
D. 数列$\{S_{n}-1\}$是等比数列

答案： 12.ACD　当$n = 1$时，$S_1 = 2a_1 + 1$，解得$a_1 = -1$，当$n\geq2$时，由$S_n = 2a_n + 1$，得$S_{n - 1} = 2a_{n - 1} + 1$，两式相减得$a_n = 2a_n - 2a_{n - 1}$，故$a_n = 2a_{n - 1}$，所以数列$\{a_n\}$是以$-1$为首项，$2$为公比的等比数列，故C正确；可得$a_n = (-1)\times2^{n - 1}$，$S_n = \frac{-1\times(1 - 2^n)}{1 - 2} = 1 - 2^n$. 令$n = 5$，得$a_5 = -16$，$S_5 = 1 - 2^5 = -31$，故A正确，B错误；因为$S_n = 1 - 2^n$，所以$S_n - 1 = -2^n$，故D正确. 故选ACD.

13.（2024·晋城期末）已知数列$\{a_{n}\}$的首项为2，且满足$a_{n + 1}=\frac{2a_{n}}{3a_{n}+1}$，则$\frac{1}{a_{n}}=$________.

答案： 13.$3 - \frac{5}{2^n}$
解析　由题意知$\frac{1}{a_{n + 1}} = \frac{3a_n + 1}{2a_n} = \frac{3}{2} + \frac{1}{2a_n}$，则$\frac{1}{a_{n + 1}} - 3 = \frac{1}{2}(\frac{1}{a_n} - 3)$，所以$\frac{\frac{1}{a_{n + 1}} - 3}{\frac{1}{a_n} - 3} = \frac{1}{2}$，因为$\frac{1}{a_1} - 3 = -\frac{5}{2}$，所以数列$\{\frac{1}{a_n} - 3\}$是以$-\frac{5}{2}$为首项，$\frac{1}{2}$为公比的等比数列，所以$\frac{1}{a_n} - 3 = (-\frac{5}{2})\times(\frac{1}{2})^{n - 1} = -5\times\frac{1}{2^n}$，则$\frac{1}{a_n} = 3 - \frac{5}{2^n}$.

14. 已知数列$\{a_{n}\}$的前$n$项和为$S_{n}$，若$\frac{S_{1}}{3}+\frac{S_{2}}{4}+\frac{S_{3}}{5}+\cdots+\frac{S_{n}}{n + 2}=n^{2}+n$.
(1)求数列$\{a_{n}\}$的通项公式；
(2)证明：$\frac{1}{S_{1}}+\frac{1}{S_{2}}+\frac{1}{S_{3}}+\cdots+\frac{1}{S_{n}}＜\frac{3}{8}$.

答案： 14.解　
(1)因为$\frac{S_1}{3} + \frac{S_2}{4} + \frac{S_3}{5} + \cdots + \frac{S_n}{n + 2} = n^2 + n$①，所以当$n\geq2$时，$\frac{S_1}{3} + \frac{S_2}{4} + \frac{S_3}{5} + \cdots + \frac{S_{n - 1}}{n + 1} = (n - 1)^2 + (n - 1)$②，①$-$②得$\frac{S_n}{n + 2} = 2n$，所以$S_n = 2n(n + 2)(n\geq2)$. 又当$n = 1$时，$S_1 = 6$也符合上式，所以$S_n = 2n(n + 2)(n\in N^*)$. 所以当$n\geq2$时，$a_n = S_n - S_{n - 1} = 2n(n + 2) - 2(n - 1)(n + 1) = 4n + 2$，当$n = 1$时，$a_1 = S_1 = 6$也符合上式，故$a_n = 4n + 2(n\in N^*)$.
(2)证明：由
(1)知$\frac{1}{S_n} = \frac{1}{2n(n + 2)} = \frac{1}{4}(\frac{1}{n} - \frac{1}{n + 2})$，所以$\frac{1}{S_1} + \frac{1}{S_2} + \frac{1}{S_3} + \cdots + \frac{1}{S_n} = \frac{1}{4}[(1 - \frac{1}{3}) + (\frac{1}{2} - \frac{1}{4}) + (\frac{1}{3} - \frac{1}{5}) + (\frac{1}{4} - \frac{1}{6}) + \cdots + (\frac{1}{n - 1} - \frac{1}{n + 1}) + (\frac{1}{n} - \frac{1}{n + 2})] = \frac{1}{4}(1 + \frac{1}{2} - \frac{1}{n + 1} - \frac{1}{n + 2}) = \frac{1}{4}(\frac{3}{2} - \frac{1}{n + 1} - \frac{1}{n + 2}) = \frac{3}{8} - \frac{1}{4}(\frac{1}{n + 1} + \frac{1}{n + 2}) \lt \frac{3}{8}$.

15.（逻辑推理）已知数列$\{a_{n}\}$满足$a_{1}=15,a_{n + 1}-a_{n}=2n(n\in N^{*})$，则$\frac{a_{n}}{n}$的最小值为_______.

答案： 15.$\frac{27}{4}$
解析　因为$a_{n + 1} - a_n = 2n$，所以$a_n - a_1 = a_n - a_{n - 1} + a_{n - 1} - a_{n - 2} + \cdots + a_2 - a_1 = n^2 - n$. 又$a_1 = 15$，所以$a_n = n^2 - n + 15$，则$\frac{a_n}{n} = n + \frac{15}{n} - 1$. 由对勾函数的单调性可知，当$n = 4$时，$\frac{a_n}{n}$取得最小值，最小值为$\frac{27}{4}$.

16.（逻辑推理）若数列$\{a_{n}\}$满足$a_{1}+2a_{2}+4a_{3}+\cdots+2^{n - 1}a_{n}=8n(n\in N^{*})$，则$a_{n}=$________.

答案： 16.$2^{4 - n}$
解析　当$n = 1$时，$a_1 = 8$. 因为$a_1 + 2a_2 + 4a_3 + \cdots + 2^{n - 1}a_n = 8n$，所以当$n\geq2$时，$a_1 + 2a_2 + 4a_3 + \cdots + 2^{n - 2}a_{n - 1} = 8n - 8$，两式相减得$2^{n - 1}a_n = 8 = 2^3$，所以$a_n = 2^{4 - n}(n\geq2)$. 当$n = 1$时，$a_1 = 8$满足$a_n = 2^{4 - n}$，所以$a_n = 2^{4 - n}$.