8.3数列递推式的处理

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数列递推式的处理

A 组习题

习\(\displaystyle \quad\)题

A组

1. 已知正项数列\(\displaystyle \{a_n\}\)前\(\displaystyle n\)项和为\(\displaystyle S_n\)，数列\(\displaystyle \{\sqrt{S_n}\}\)前\(\displaystyle n\)项和为\(\displaystyle G_n\)，且\(\displaystyle S_n+\frac{1}{2}G_n=1\)，求\(\displaystyle a_4\)。

2. 已知等比数列\(\displaystyle \{a_n\}\)的前\(\displaystyle n\)项和为\(\displaystyle S_n\),\(\displaystyle a_1>1\),\(\displaystyle S_3=\mathrm{e}^{S_1}\),则数列\(\displaystyle \{a_n\}\)的公比\(\displaystyle q\)满足

   • \(\displaystyle 0<q\leqslant1\)
   • \(\displaystyle -1<q\leqslant0\)
   • \(\displaystyle q>1\)
   • \(\displaystyle q\leqslant-1\)

   3. 已知数列\(\displaystyle \{a_n\}\)满足\(\displaystyle a_1=1\)，\(\displaystyle a_{n+1}=\frac{a_n}{a_n^2+1}\)（\(\displaystyle n\in\mathbb{N}^*\)），记\(\displaystyle \{a_n\}\)的前\(\displaystyle n\)项和为\(\displaystyle S_n\)，证明：\(\displaystyle a_{2026}(S_{2025}+1)=1\) 4. 【2025“集英苑测试”（网络联考）10】（多选）已知数列\(\displaystyle \left \{ a_n \right \}\)满足\(\displaystyle a_{n+1}\geqslant a_n,a_na_{n+1}a_{n+2}=8^n(n\in\mathbb{N^*})\)，则
   • \(\displaystyle a_{n+3}=8a_n\)
   • \(\displaystyle a_n>0\)
   • \(\displaystyle a_3^2a_4\leqslant 64\)
   • \(\displaystyle a_1\in[\frac{1}{2},2]\)

   5. （多选）已知无穷数列\(\displaystyle \{a_n\}\)的前\(\displaystyle n\)项和为\(\displaystyle S_n\),则下列说法正确的是
   • 若\(\displaystyle S_n-a_n\geqslant0\),则\(\displaystyle a_n\geqslant0\)
   • 若\(\displaystyle S_n-a_n\geqslant0\),则\(\displaystyle S_n\geqslant0\)
   • 若\(\displaystyle S_n+a_n\geqslant0\),则\(\displaystyle S_n\geqslant a_2\)
   • 若\(\displaystyle S_n+a_n\geqslant0\),则\(\displaystyle S_2\geqslant a_4\)

   6. （多选）已知数列\(\displaystyle \{a_n\}\)满足\(\displaystyle a_{n}+a_{n+2}=\lambda a_{n+1}\),\(\displaystyle \lambda\in\vv{R}\),若\(\displaystyle a_1=1\),\(\displaystyle a_2=2\),\(\displaystyle a_{2024}=2024\),则\(\displaystyle \lambda\)的值可能为
   • \(\displaystyle -1\)
   • \(\displaystyle 2\)
   • \(\displaystyle \frac{5}{2}\)
   • \(\displaystyle -2\)

   7. （多选）称递推公式\(\displaystyle \Psi\)是专一的，若无穷数列\(\displaystyle \{a_n\}\)由\(\displaystyle \Psi\)唯一确定，则下列递推公式中专一的有
   • \(\displaystyle \{(a_n,a_{n+1})=\{2n-1,2n+1\},n\in\mathbb{N}^*\}\)
   • \(\displaystyle \begin{cases}a_1=a_2=1,\\(a_{n+2}-a_{n+1})^2=a_n,n\in\mathbb{N}^*\end{cases}\)
   • \(\displaystyle \begin{cases}a_2=2,\\a_{n+1}a_n=a_n+a_{n+1},n\in\mathbb{N}^*\end{cases}\)
   • \(\displaystyle \begin{cases}a_1+a_{n+1}=2n,\\a_n a_{n+2}=3n,n\in\mathbb{N}^*\end{cases}\)

   8. 【2022北京15】（多选）已知正项数列\(\displaystyle \left \{ a_n \right \}\)的前\(\displaystyle n\)项和\(\displaystyle S_n\)满足\(\displaystyle a_n\cdot S_n=9(n=1,2,\cdots)\)，下列命题中正确的是
   • \(\displaystyle a_2<3\)
   • \(\displaystyle \left \{ a_n \right \}\)为等比数列
   • \(\displaystyle \left \{ a_n \right \}\)为递减数列
   • \(\displaystyle \left \{ a_n \right \}\)中存在小于\(\displaystyle \frac{1}{100}\)的项
   答案

   新答案（来源：1.29 数列综合.md）： ①③④

   【解题思路】如下表．

   \[\displaystyle \begin{array}{c|c|l} \text{选项} & \text{判断} & \text{解析}\\ \hline \text{①} & \text{正确} & a_1S_1=a_1^2=9\text{，则}a_1=3\text{．而}a_2S_2=a_2\left(a_1+a_2\right)=9\text{，即}a_2^2=9-a_1a_2<9\text{，所以}a_2<3\text{．}\\ \text{②} & \text{错误} & \text{反证法．若}\left\{a_n\right\}\text{是等比数列，则}a_n=3q^{n-1}\text{，其中}q\text{是公比，}S_n=\frac{3\left(1-q^n\right)}{1-q}\text{．故}a_nS_n=9\text{可以化为}q^{n-1}\left(1-q^n\right)=1-q\text{对任意正整数}n\text{恒成立，但这是显然不成立的．}\\ \text{③} & \text{正确} & a_{n+1}-a_n=\frac{9}{S_{n+1}}-\frac{9}{S_n}=\frac{9\left(S_n-S_{n+1}\right)}{S_nS_{n+1}}=-\frac{9a_{n+1}}{S_nS_{n+1}}<0\text{．}\\ \text{④} & \text{正确} & \text{反证法．若}\left\{a_n\right\}\text{每项都大于等于}\frac{1}{100}\text{，则}S_n\geqslant\frac{n}{100}\text{，于是}9=a_nS_n\geqslant\frac{1}{100}\cdot\frac{n}{100}\text{对任意正整数}n\text{恒成立．然而当}n>90000\text{时上述不等式不成立，矛盾．} \end{array}\]

   本题有错得\(\displaystyle 0\)分，少选\(\displaystyle 1\)个得\(\displaystyle 4\)分，少选\(\displaystyle 2\)个得\(\displaystyle 3\)分．

   【实测数据】本题平均分\(\displaystyle 2.94\)分，标准差\(\displaystyle 1.28\)，得分率\(\displaystyle 0.588\)，鉴别指数\(\displaystyle 0.31\)．

   \[\displaystyle \begin{array}{c|ccccc} \text{得分} & 0 & 3 & 4 & 5\\ \hline \text{比率}(\%) & 13.42 & 57.49 & 23.96 & 5.12 \end{array}\]

   \(\displaystyle G_1\)至\(\displaystyle G_{10}\)组得分率如下：

   \[\displaystyle \begin{array}{c|cccccccccc} & G_1 & G_2 & G_3 & G_4 & G_5 & G_6 & G_7 & G_8 & G_9 & G_{10}\\ \hline \text{得分率} & 0.37 & 0.45 & 0.50 & 0.54 & 0.57 & 0.61 & 0.64 & 0.67 & 0.72 & 0.80 \end{array}\]

   1. 【2007广东理21】 已知函数 \(\displaystyle f(x) = x^2 + x - 1\), \(\displaystyle \alpha,\beta\) 是方程 \(\displaystyle f(x)=0\) 的两个根 (\(\displaystyle \alpha>\beta\)), \(\displaystyle f'(x)\) 是 \(\displaystyle f(x)\) 的导数, 设 \(\displaystyle a_1=1\), \(\displaystyle a_{n+1} = a_n - \frac{f(a_n)}{f'(a_n)}\) (\(\displaystyle n=1,2,\cdots\))。

   2. 证明: 对任意的正整数 \(\displaystyle n\), 都有 \(\displaystyle a_n > \alpha\)；

   3. 记 \(\displaystyle b_n = \ln \frac{a_n - \beta}{a_n - \alpha}\) (\(\displaystyle n=1,2,\cdots\)), 求数列 \(\displaystyle \{b_n\}\) 的前 \(\displaystyle n\) 项和 \(\displaystyle S_n\)。
   4. 【2006天津文21】已知数列 \(\displaystyle \{x_n\}\) 满足 \(\displaystyle x_1 = x_2 = 1\), 并且 \(\displaystyle \frac{x_{n+1}}{x_n} = \lambda \frac{x_n}{x_{n-1}}\) (\(\displaystyle \lambda\) 为非零参数, \(\displaystyle n=2,3,4,\cdots\))。

   （1） 若 \(\displaystyle x_1, x_3, x_5\) 成等比数列, 求参数 \(\displaystyle \lambda\) 的值；

   （2） 设 \(\displaystyle 0 < \lambda < 1\), 常数 \(\displaystyle k \in \mathbb{N}^*\) 且 \(\displaystyle k \geqslant 3\), 证明: \(\displaystyle \frac{x_{1+k}}{x_1} + \frac{x_{2+k}}{x_2} + \cdots + \frac{x_{n+k}}{x_n} < \frac{\lambda^k}{1-\lambda^k}\) (\(\displaystyle n \in \mathbb{N}^*\))。 11. 【2019海南新高考适应性测试22】已知数列\(\displaystyle \left \{ a_n\right \}\)的首项不为零，其前\(\displaystyle n\)项和为\(\displaystyle S_n\)，满足\(\displaystyle 2\sqrt{S_n}=a_n+c\)。

   （1）证明：\(\displaystyle c\leqslant 1\)；

   （2）若\(\displaystyle c=0\)，证明：\(\displaystyle S_n\geqslant (n+1)^2\)；

   （3）是否存在常数\(\displaystyle c\)，使得\(\displaystyle \left \{ a_n \right \}\)是等比数列？若存在，求出\(\displaystyle c\)的所有可能值，否则请说明理由。 12. 【2026成都二诊19】在数列\(\displaystyle \{a_n\}\)中，\(\displaystyle a_1=1\),\(\displaystyle a_{n+1}=\sqrt{a_n+2}\)，设\(\displaystyle \{a_n\}\)的前\(\displaystyle n\)项和为\(\displaystyle S_n\)。

   1. 是否存在常数$\displaystyle A,\omega(A>0,0<\omega<2\pi)$,使得$\displaystyle a_n=A\cos\frac{\omega}{2^n}$？若存在，求$\displaystyle A$和$\displaystyle \omega$的值，否则请说明理由；
   

   1. 求\(\displaystyle [2S_1+1]+[2S_2+1]+\dots+[2S_n+1]\).

B 组习题

B组

1. 【2023 北京10】

   已知数列\(\displaystyle \left\{a_n\right\}\)满足\(\displaystyle a_{n+1}=\frac{1}{4}\left(a_n-6\right)^3+6\left(n=1,2,3,\cdots\right)\)，则

   • \(\displaystyle 当a_1=3时，\{a_n\}为递减数列，且存在常数M\leqslant0，使得a_n>M恒成立\)
   • \(\displaystyle 当a_1=5时，\{a_n\}为递增数列，且存在常数M\leqslant6，使得a_n<M恒成立\)
   • \(\displaystyle 当a_1=7时，\{a_n\}为递减数列，且存在常数M>6，使得a_n>M恒成立\)
   • \(\displaystyle 当a_1=9时，\{a_n\}为递增数列，且存在常数M>0，使得a_n<M恒成立\)
   答案

   B. 因为\(\displaystyle a_{n+1}-6=\frac{1}{4}\left(a_n-6\right)^3\)，当\(\displaystyle a_1<6\)时，\(\displaystyle a_n-6<0\)恒成立；当\(\displaystyle a_1>6\)时，\(\displaystyle a_n-6>0\)恒成立；故只要\(\displaystyle a_1\ne6\)，那么\(\displaystyle a_n\ne6\)，此时取以\(\displaystyle 2\)为底的对数可得 $\(\displaystyle \log_2\left|a_{n+1}-6\right|=3\log_2\left|a_n-6\right|-2,\Longleftrightarrow\log_2\left|a_{n+1}-6\right|-1=3\left(\log_2\left|a_n-6\right|-1\right).\)$ 因此\(\displaystyle \left\{\log_2\left|a_{n+1}-6\right|-1\right\}\)是公比为\(\displaystyle 3\)的等比数列，从而\(\displaystyle \log_2\left|a_n-6\right|-1=\left(\log_2\left|a_1-6\right|-1\right)\cdot3^{n-1}\)，即 $\(\displaystyle \left|a_n-6\right|=2\times2^{\left(\log_2\left|a_1-6\right|-1\right)\cdot3^{n-1}}=2\times\left(\frac{1}{2\left|a_1-6\right|}\right)^{3^{n-1}}.\)$ ①当\(\displaystyle a_1=3\)时，\(\displaystyle a_n=6-2\times\left(\frac{3}{2}\right)^{3^{n-1}}\)，故\(\displaystyle \left\{a_n\right\}\)为递减数列，但是\(\displaystyle a_n\to-\infty\left(n\to\infty\right)\)，\(\displaystyle \left\{a_n\right\}\)没有下界，故不存在常数\(\displaystyle M\leqslant0\)使得\(\displaystyle a_n>M\)恒成立，故\(\displaystyle A\)错误.

   ②当\(\displaystyle a_1=5\)时，\(\displaystyle a_n=6-2\times\left(\frac{1}{2}\right)^{3^{n-1}}\)，故\(\displaystyle \left\{a_n\right\}\)为递增数列，并且对常数\(\displaystyle M=6\)，有\(\displaystyle a_n<6\)恒成立，故\(\displaystyle B\)正确.

   ③当\(\displaystyle a_1=7\)时，\(\displaystyle a_n=6+2\times\left(\frac{1}{2}\right)^{3^{n-1}}\)，故\(\displaystyle \left\{a_n\right\}\)为递减数列，但是\(\displaystyle a_n\to6\left(n\to\infty\right)\)，故不存在大于\(\displaystyle 6\)的常数\(\displaystyle M\)使得\(\displaystyle a_n>M\)恒成立，故\(\displaystyle C\)错误．

   ④当\(\displaystyle a_1=9\)时，\(\displaystyle a_n=6+2\times\left(\frac{3}{2}\right)^{3^{n-1}}\)，故\(\displaystyle \left\{a_n\right\}\)为递增数列，但是\(\displaystyle a_n\to+\infty\left(n\to\infty\right)\)，\(\displaystyle \left\{a_n\right\}\)没有上界，故不存在常数\(\displaystyle M>0\)使得\(\displaystyle a_n<M\)恒成立，故\(\displaystyle D\)错误.

   【实测数据】本题难度\(\displaystyle 0.52\)，标准差\(\displaystyle 2.00\)，区分度\(\displaystyle 0.46\)．得分率较低，说明考生的逻辑推理素养水平不乐观．

   1. 【2012 四川16改编】（多选）设\(\displaystyle a\)为正整数，数列\(\displaystyle \left\{x_n\right\}\)满足\(\displaystyle x_1=a\)，\(\displaystyle x_{n+1}=\left[\frac{x_n+\left[\frac{a}{x_n}\right]}{2}\right]\left(n\in\mathbb{N}^{*}\right)\)．则下列命题正确的是

      <div class="choices choices--4" markdown>
      

   2. 当\(\displaystyle a=5\)时，数列\(\displaystyle \left\{x_n\right\}\)的前\(\displaystyle 3\)项依次为\(\displaystyle 5,3,2\)

   3. 对数列\(\displaystyle \left\{x_n\right\}\)，都存在正整数\(\displaystyle k\)，当\(\displaystyle n\geqslant k\)时总有\(\displaystyle x_n=x_k\)
   4. 当\(\displaystyle n\geqslant1\)时，\(\displaystyle x_n>\sqrt{a}-1\)
   5. 对某个正整数\(\displaystyle k\)，若\(\displaystyle x_{k+1}\geqslant x_k\)，则\(\displaystyle x_k=\left[\sqrt{a}\right]\)