《2025年绿色通道45分钟课时作业与单元测评高中数学选择性必修第二册人教版》

11. 已知数列$\{ a_{n}\} $满足$a_{n+1}=\frac {2}{3}a_{n}+4$,且$a_{1}=1$,则$\{ a_{n}\} $的通项公式为(C)

A.$a_{n}=12-\left(\frac {2}{3}\right)^{n-1}$
B.$a_{n}=\left(\frac {2}{3}\right)^{n+2}$
C.$a_{n}=12-11× \left(\frac {2}{3}\right)^{n-1}$
D.$a_{n}=8+\left(\frac {2}{3}\right)^{n-1}$

答案： 11.C 设$a_{n+1}+x=\frac{2}{3}(a_n+x)$，即$a_{n+1}=\frac{2}{3}a_n-\frac{1}{3}x$，所以$-\frac{1}{3}x=4$，解得$x=-12$，所以$a_{n+1}-12=\frac{2}{3}(a_n-12)$，所以$\{a_n-12\}$是首项为$a_1-12=-11$，公比为$\frac{2}{3}$的等比数列，所以$a_n-12=-11×(\frac{2}{3})^{n-1}$，所以$a_n=12-11×(\frac{2}{3})^{n-1}$.

12. (2024·河南许昌期中)数列$\{ a_{n}\} $的首项$a_{1}=2$,且$a_{n+1}=4a_{n}+6(n\in \boldsymbol {N}^{*})$,令$b_{n}=\log$${2}(a_{n}+2)$,则$\frac {b_{1}+b_{2}+·s +b_{2\ 021}}{2\ 021}=$(C)

A.$2\ 020$
B.$2\ 021$
C.$2\ 022$
D.$2\ 023$

答案： 12.C $\because a_{n+1}=4a_n+6(n\in N^{*}), \therefore a_{n+1}+2=4a_n+6+2=4(a_n+2)＞0$，即$\frac{a_{n+1}+2}{a_n+2}=4$且$a_1=2, \therefore$数列$\{a_n+2\}$是以$4$为首项，$4$为公比的等比数列，故$a_n+2=4^n$，由$b_n=\log_2(a_n+2)$，得$b_n=\log_2 4^n=2n$.设数列$\{b_n\}$的前$n$项和为$S_n$，则$S_{2021}=2×(1+2+3+·s+2020+2021)=2021× 2021+1$，$\therefore \frac{b_1+b_2+·s+b_{2021}}{2021}=\frac{2021×(1+2021)}{2021}=2022$.

13. 已知数列$\{ a_{n}\} $的前$n$项之和为$S_{n}$,$a_{1}=\frac {2}{3}$,$a_{n+1}=\frac {(2n+4)a_{n}}{(n^{2}+5n+6)a_{n}+2n+6}$,则$S_{9}=$(D)

A.$\frac {10}{11}$
B.$\frac {1}{11}$
C.$\frac {82}{55}$
D.$\frac {72}{55}$

答案： 13.D $a_{n+1}=\frac{(2n+4)a_n}{(n^2+5n+6)a_n+2n+6}=\frac{2(n+2)a_n}{(n+2)(n+3)a_n+2(n+3)}=\frac{2(n+2)a_n}{(n+3)[(n+2)a_n+2]}$，则$\frac{1}{(n+3)a_{n+1}}=\frac{(n+2)a_n+2}{2(n+2)a_n}$，取倒数有$\frac{1}{(n+2)a_n}-\frac{1}{(n+3)a_{n+1}}=\frac{1}{2}$，则数列$\{\frac{1}{(n+2)a_n}\}$是以$\frac{1}{3a_1}=\frac{1}{2}$为首项，$\frac{1}{2}$为公差的等差数列，则$\frac{1}{(n+2)a_n}=\frac{1}{2}+\frac{1}{2}(n-1)=\frac{n}{2}$，则$a_n=\frac{2}{n(n+2)}=\frac{1}{n}-\frac{1}{n+2}$，则$S_9=a_1+a_2+·s+a_9=1-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+·s+\frac{1}{9}-\frac{1}{11}=1+\frac{1}{2}-\frac{1}{10}-\frac{1}{11}=\frac{72}{55}$，故选D.

14. 在数列$\{ a_{n}\} $中,$a_{1}=2$,$a_{2}=8$,且对任意的$n\in \boldsymbol {N}^{*}$,都有$a_{n+2}=4a_{n+1}-4a_{n}$.
(1) 证明$\{ a_{n+1}-2a_{n}\} $是等比数列,并求出$\{ a_{n}\} $的通项公式;
(2) 若$b_{n}=\begin{cases}\dfrac {a_{n}}{n},n=2k-1,\\\log \underline{\quad\quad} {2}\dfrac {a_{n}}{n· 2^{n+1}},n=2k\end{cases}k\in \boldsymbol {N}^{*}$,且数列$\{ b_{n}\} $的前$n$项积为$T_{n}$,求$T_{15}$和$T_{20}$.

答案： 14.解
(1)由题意可得$a_{n+2}-2a_{n+1}=2(a_{n+1}-2a_n), a_2-2a_1=4$，所以$\{a_{n+1}-2a_n\}$是以$4$为首项，$2$为公比的等比数列，所以$a_{n+1}-2a_n=4× 2^{n-1}=2^{n+1} \Rightarrow \frac{a_{n+1}}{2^{n+1}}-\frac{a_n}{2^n}=1$，所以$\{\frac{a_n}{2^n}\}$是以$1$为首项，$1$为公差的等差数列，则$\frac{a_n}{2^n}=n, a_n=n· 2^n$.
(2)由
(1)可得$b_n=\begin{cases} 2^n, n=2k-1, \\ -1, n=2k, \end{cases} k\in N^{*}$，所以$T_{15}=b_1× b_2× ·s× b_{15}=2^1× 2^3× ·s× 2^{15}× (-1)^{7}=-2^{64}$；同理$T_{20}=2^1× 2^3× ·s× 2^{19}× (-1)^{10}=2^{100}$.

15. 数列$\{ a_{n}\} $满足$a_{1}=\frac {1}{2}$,$a_{n+1}-a_{n}=p· 3^{n-1}-nq$.
(1) 若$q=0$,且数列$\{ a_{n}\} $为等比数列,求$p$的值;
(2) 若$p=1$,且$a_{4}$为数列$\{ a_{n}\} $的最小项,求$q$的取值范围.

答案： 15.解
(1)当$q=0$时，$a_{n+1}-a_n=p· 3^{n-1}$，$\because$数列$\{a_n\}$为等比数列，$\therefore$令$a_{n+1}=t a_n$，$\therefore \begin{cases} a_2-a_1=p· 3^{0}=p, \\ a_3-a_2=p· 3^{1}=3p, \end{cases}$又$\because a_3-a_2=t(a_2-a_1)$，$\therefore t=3, \therefore a_2-a_1=2a_1=1=p$.
(2)若$p=1$，则$a_{n+1}-a_n=3^{n-1}-nq$，$\therefore a_n=a_1+(a_2-a_1)+(a_3-a_2)+·s+(a_n-a_{n-1})=\frac{1}{2}+(1+3+·s+3^{n-2})-[1+2+·s+(n-1)]q=\frac{1}{2}[3^{n-1}-n(n-1)q]$.$\because$数列$\{a_n\}$的最小项为$a_4, \therefore$对$\forall n\in N^{*}$，有$\frac{1}{2}[3^{n-1}-n(n-1)q]\geq a_4=\frac{1}{2}(27-12q)$恒成立，即$3^{n-1}-27\geq (n^2-n-12)q$对$\forall n\in N^{*}$成立.当$n=1$时，有$-26\geq -12q, \therefore q\geq \frac{13}{6}$；当$n=2$时，有$-24\geq -10q, \therefore q\geq \frac{12}{5}$；当$n=3$时，有$-18\geq -6q, \therefore q\geq 3$；当$n=4$时，有$0\geq 0, \therefore q\in R$；当$n\geq 5$时，$n^2-n-12＞0, \therefore$有$q\leq \frac{3^{n-1}-27}{n^2-n-12}$恒成立，令$c_n=\frac{3^{n-1}-27}{n^2-n-12}(n\geq 5,n\in N^{*})$，则$c_{n+1}-c_n=\frac{2(n^2-2n-12)3^{n-1}+54n}{(n^2-16)(n^2-9)}＞0$，即数列$\{c_n\}$为递增数列，$\therefore q\leq c_5=\frac{27}{4}$.综上所述，$3\leq q\leq \frac{27}{4}$.