《2026年南方凤凰台5A新考案数学二轮提高版》

例 1 - 2 设数列$\{ a_{n}\}$的前$n$项积为$T_{n}$，满足$a_{n} + 2T_{n} = 1(n\in\mathbf{N}^{*})$，则$T_{n} =$1/(2n+1).

答案： 例1-2$\frac{1}{2n+1}$【解析】因为$a_{n}+2T_{n}=1(n\in N^{*})$,所以$\frac{T_{n}}{T_{n-1}}+2T_{n}=1(n\geqslant2,n\in N^{*})$,所以$T_{n}+2T_{n}T_{n-1}=T_{n-1}$,所以$\frac{1}{T_{n}}-\frac{1}{T_{n-1}}=2$,所以$\{\frac{1}{T_{n}}\}$是公差为$2$的等差数列.因为$a_{1}+2T_{1}=1$,即$3a_{1}=1$,所以$a_{1}=T_{1}=\frac{1}{3}$,所以$\frac{1}{T_{n}}=\frac{1}{3}+(n-1)×2=2n+1$,$T_{n}=\frac{1}{2n+1}$.

例 1 - 3 已知数列$\{ a_{n}\}$满足$a_{1} + \frac{1}{2}a_{2} + \frac{1}{2^{2}}a_{3} + ·s + \frac{1}{2^{n - 1}}a_{n} = n$，$n\in\mathbf{N}^{*}$，则数列$\{ a_{n}\}$的通项公式为$a_{n} =$2ⁿ⁻¹.

答案： 例1-3$2^{n-1}$【解析】由$a_{1}+\frac{1}{2}a_{2}+\frac{1}{2^{2}}a_{3}+·s+\frac{1}{2^{n-1}}a_{n}=n$①,得当$n\geqslant2$时,$a_{1}+\frac{1}{2}a_{2}+\frac{1}{2^{2}}a_{3}+·s+\frac{1}{2^{n-2}}a_{n-1}=n-1$②,由①-②,可得$\frac{1}{2^{n-1}}a_{n}=1$,所以$a_{n}=2^{n-1}$.当$n=1$时,$a_{1}=1$,满足上式,所以数列$\{a_{n}\}$的通项公式为$a_{n}=2^{n-1}$.

变式 1 已知正项数列$\{ a_{n}\}$的前$n$项和为$S_{n}$，且$6S_{n} = (a_{n} + 7)(a_{n} - 4)$，数列$\{ b_{n}\}$满足$b_{1} = 4$，$3b_{1} + 3b_{2} + 3b_{3} + ·s + 3b_{n} = 4^{n + 1} - 4$，求数列$\{ a_{n}\}$和$\{ b_{n}\}$的通项公式.{aₙ}的通项公式为aₙ=3n+4，{bₙ}的通项公式为bₙ=4ⁿ

答案： 变式1【解答】
(1)由$6S_{n}=(a_{n}+7)(a_{n}-4)=a_{n}^{2}+3a_{n}-28$,得$6S_{n+1}=a_{n+1}^{2}+3a_{n+1}-28$,两式相减得$6a_{n+1}=a_{n+1}^{2}-a_{n}^{2}+3a_{n+1}-3a_{n}$,即$(a_{n+1}+a_{n})(a_{n+1}-a_{n}-3)=0$.因为$a_{n}＞0$,所以$a_{n+1}-a_{n}-3=0$,即$a_{n+1}-a_{n}=3$.当$n=1$时,$6a_{1}=a_{1}^{2}+3a_{1}-28$,解得$a_{1}=7$或$a_{1}=-4$(舍去),所以$\{a_{n}\}$是首项为$7$,公差为$3$的等差数列,故$a_{n}=3n+4$.因为$3b_{1}+3b_{2}+3b_{3}+·s+3b_{n}=4^{n+1}-4$①,所以当$n\geqslant2$时,$3b_{1}+3b_{2}+3b_{3}+·s+3b_{n-1}=4^{n}-4$②,①-②得$b_{n}$.当$n=1$时,可得$b_{1}=4$,也满足$b_{n}=4^{n}$.故$\{a_{n}\}$的通项公式为$a_{n}=3n+4$,$\{b_{n}\}$的通项公式为$b_{n}=4^{n}$.

例 2 (1)已知数列$\{ a_{n}\}$满足$a_{1} = \frac{1}{2}$，且对任意的$n\in\mathbf{N}^{*}$，都有$a_{n + 1} = \frac{a_{n}}{2} + \frac{n}{2} + 1$，则数列$\{ a_{n}\}$的通项公式为aₙ=-(1/2)ⁿ+n.

答案： 例2
(1)$a_{n}=-(\frac{1}{2})^{n}+n$【解析】因为$a_{n+1}=\frac{a_{n}}{2}+\frac{n}{2}+1$,所以设$a_{n+1}+k(n+1)+b=\frac{1}{2}(a_{n}+kn+b)$,所以$a_{n+1}=\frac{1}{2}a_{n}-\frac{kn}{2}-k-\frac{b}{2}$,$\begin{cases}k=-1\\-k-\frac{b}{2}=1\end{cases}$,解得$\begin{cases}k=-1\\b=0\end{cases}$,1)$=\frac{1}{2}(a_{n}-n)$.设$b_{n}=a_{n}-n$,则$b_{n+1}=\frac{1}{2}b_{n}$.因为$a_{1}=\frac{1}{2}$,所以$b_{1}=\frac{1}{2}-1=-\frac{1}{2}$,所以数列$\{b_{n}\}$是首项为$-\frac{1}{2}$,公比为$\frac{1}{2}$的等比数列,所以$b_{n}=-\frac{1}{2}·(\frac{1}{2})^{n-1}=-(\frac{1}{2})^{n}$,所以$a_{n}=b_{n}+n=-(\frac{1}{2})^{n}+n$.

(2)(2025·临沂 2 月模考)设数列$\{ a_{n}\}$的前$n$项和为$S_{n}$，且$S_{n} + na_{n} = 1$，则满足$S_{n} ＞ 0.99$时，$n$的最小值为(D)

A.49
B.50
C.99
D.100

答案： 例2
(2)D【解析】因为$S_{n}+na_{n}=1$,所以$a_{1}=\frac{1}{2}$,当$n\geqslant2$时,$S_{n}+na_{n}=S_{n-1}+(n-1)a_{n-1}=1$,所以$(n+1)a_{n}=(n-1)a_{n-1}$,方法一:即$\frac{a_{n}}{a_{n-1}}=\frac{n-1}{n+1}$,此时$a_{n}=\frac{a_{n}}{a_{n-1}}×\frac{a_{n-1}}{a_{n-2}}×\frac{a_{n-2}}{a_{n-3}}×·s×\frac{a_{2}}{a_{1}}× a_{1}=\frac{n-1}{n+1}×\frac{n-2}{n}×\frac{n-3}{n-1}×·s×\frac{3}{5}×\frac{2}{4}×\frac{1}{3}×\frac{1}{2}=\frac{\frac{2}{n(n+1)}×\frac{1}{2}}{1}=\frac{1}{n(n+1)}$;当$n=1$时,$a_{1}=\frac{1}{2}$也满足该式,故$a_{n}=\frac{1}{n(n+1)}$,$S_{n}=1-na_{n}=1-\frac{1}{n+1}$.令$S_{n}=\frac{1}{n+1}＞0.99$,解得$n＞99$,故所求$n$的最小值为$100$.
方法二:当$n\geqslant2$时,$(n+1)na_{n}=n(n-1)a_{n-1}=2×(2-1)a_{1}=1$,即$\{n(n+1)a_{n}\}$为常数列,则$a_{n}=\frac{1}{(n+1)n}$,$S_{n}=1-na_{n}=1-\frac{1}{n+1}$.若$S_{n}=\frac{1}{n+1}＞0.99$,解得$n＞99$,故所求$n$的最小值为$100$.

（3）（2025·安庆二模）若数列$\{ a_{n}\}$满足$a_{1} = 1$，$a_{n + 1} = a_{n}^{2} + 2a_{n}$，则使得$\sqrt{a_{n} + 1} ＞ 2025$的正整数$n$的最小值为______.

答案： 例2
(3)$6$【解析】因为$a_{n+1}=a_{n}^{2}+2a_{n}$,所以$a_{n+1}+1=(a_{n}+1)^{2}$,则$\ln(a_{n+1}+1)=2\ln(a_{n}+1)$,又$a_{1}+1=2$,所以$\{\ln(a_{n}+1)\}$是以$\ln2$为首项,$2$为公比的等比数列,所以$\ln(a_{n}+1)=2^{n-1}\ln2$,所以$a_{n}=2^{2^{n-1}}-1$,则由$\sqrt{a_{n}+1}=\sqrt{2^{2^{n-1}}}＞2025$,计算得正整数$n$的最小值为$6$.