8.1数列的基本概念
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讲义正文
数列的基本概念
等差数列与等比数列
A 组习题
习\(\displaystyle \quad\)题
A组
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【2016浙江学考14】若\(\displaystyle \left \{ a_n \right \}\)是无穷等比数列，则下列数列可能不是等比数列的是
2. 【2014辽宁8】在等差数列\(\displaystyle \left \{ a_n\right \}\)中，若数列\(\displaystyle \left \{ 2^{a_1a_n} \right \}\)是严格递减数列，则- \(\displaystyle \left \{ a_{2n} \right \}\)
- \(\displaystyle \left \{ a_{2n-1} \right \}\)
- \(\displaystyle \left \{ a_na_{n+1} \right \}\)
- \(\displaystyle \left \{ a_n +a_{n+1}\right \}\)
3. 【2020全国II卷理4】 北京天坛的圜丘坛为古代祭天的场所，分上、中、下三层。上层中心有一块圆形石板（称为天心石），环绕天心石砌 9 块扇面形石板构成第一环，向外每环依次增加 9 块。下一层的第一环比上一层的最后一环多 9 块，向外每环依次也增加 9 块。已知每层环数相同，且下层比中层多 729 块，则三层共有多少块扇面形石板(不含天心石)？ 4. 【2015福建理8文16】若\(\displaystyle a,b\)是函数\(\displaystyle f(x)=x^2-px+q(p,q>0)\)的两个不同的零点，且\(\displaystyle a,b,-2\)这三个数经过适当排序后可构成等差数列与等比数列，求\(\displaystyle p+q\)。 5. 【2012上海理18】设\(\displaystyle a_n=\frac{1}{n}\sin\frac{n\pi}{25},S_n=a_1+a_2+\cdots a_n\)，求\(\displaystyle S_1,S_2,\cdots ,S_{100}\)中正数的个数。 6. 【2005湖南文5】数列\(\displaystyle \left \{ a_n \right \}\)满足\(\displaystyle a_1=0,a_{n+1}=\frac{a_n-\sqrt{3}}{\sqrt{3}a_n+1}\)，求\(\displaystyle a_{20}\)。- \(\displaystyle d<0\)
- \(\displaystyle d>0\)
- \(\displaystyle a_1d<0\)
- \(\displaystyle a_1d>0\)
答案
新答案（来源：1.26 数列的概念.md）： B
【解题思路】思路1：由\(\displaystyle a_{n+1}=\frac{a_n-\sqrt{3}}{\sqrt{3}a_n+1}\)及\(\displaystyle a_1=0\)，求得\(\displaystyle a_2=-\sqrt{3}\)，\(\displaystyle a_3=\frac{-2\sqrt{3}}{-2}=\sqrt{3}\)，\(\displaystyle a_4=0\)．
猜想\(\displaystyle a_{n+3}=a_n\)，\(\displaystyle n\in\mathbb{N}^{*}\)，亦即\(\displaystyle a_{3n+r}=a_r\left(r=1,2\right)\)．故\(\displaystyle a_{20}=a_{3\times6+2}=a_2=-\sqrt{3}\)．
思路2：令\(\displaystyle a_n=\tan b_n\)，则
\[\displaystyle \tan b_{n+1}=\frac{\tan b_n-\tan\frac{\pi}{3}}{1+\tan\frac{\pi}{3}\cdot\tan b_n}=\tan\left(b_n-\frac{\pi}{3}\right)\]所以\(\displaystyle b_{n+1}=b_n-\frac{\pi}{3}+k_n\pi\left(k_n\in\mathbb{Z}\right)\)．
因为\(\displaystyle a_1=0\)，\(\displaystyle b_1=0\)，所以\(\displaystyle b_n=-\frac{\left(n-1\right)\pi}{3}+m_n\pi\left(m_n\in\mathbb{Z}\right)\)，其中\(\displaystyle m_n=k_1+\cdots+k_{n-1}\)（如果\(\displaystyle n=1\)，则\(\displaystyle m_1=0\)）．这样，\(\displaystyle a_{20}=\tan b_{20}=-\tan\frac{19\pi}{3}=-\tan\frac{\pi}{3}=-\sqrt{3}\)．
【实测数据】本题文科难度\(\displaystyle 0.666\)
新答案（来源：1.27 等差数列.md）：B
【解题思路】设等差数列\(\displaystyle \left\{a_n\right\}\)的公差为\(\displaystyle \frac{2\pi}{3}\)，即\(\displaystyle a_{n+3k}=a_n+3k\times\frac{2\pi}{3}=a_n+2k\pi\)，\(\displaystyle \cos a_{n+3k}=\cos a_n\)，这样集合
\[\displaystyle S=\left\{\cos a_n\left|n\in\mathbb{N}^{*}\right.\right\}=\left\{\cos a_n\left|n=1,2,3\right.\right\}=\left\{a,b\right\}\]因此\(\displaystyle a_1\)，\(\displaystyle a_2\)，\(\displaystyle a_3\)中有两个数的余弦值相等．
思路1：不妨设\(\displaystyle \cos a_1=\cos a_2\)，满足\(\displaystyle y=\cos x=\cos\left(x+\frac{2\pi}{3}\right)\)的\(\displaystyle y\)只有\(\displaystyle y=\frac{1}{2}=\cos\left(-\frac{\pi}{3}\right)=\cos\left(\frac{\pi}{3}\right)\)或\(\displaystyle y=-\frac{1}{2}=\cos\left(\frac{2\pi}{3}\right)=\cos\left(\frac{4\pi}{3}\right)\)，两种情况对应第三个数的余弦值分别为\(\displaystyle \cos\pi=-1\)或\(\displaystyle \cos2\pi=1\)，因此乘积\(\displaystyle ab=\frac{1}{2}\times\left(-1\right)\)或\(\displaystyle ab=\left(-\frac{1}{2}\right)\times1\)，均有\(\displaystyle ab=-\frac{1}{2}\)，故正确选项为\(\displaystyle B\)．
思路2：将等差数列的项\(\displaystyle a_n\)对应为单位圆周上的点\(\displaystyle \left(\cos a_n,\sin a_n\right)\)，由于\(\displaystyle \left\{a_n\right\}\)的公差是\(\displaystyle \frac{2\pi}{3}\)，即\(\displaystyle a_{n+3k}\)与\(\displaystyle a_n\)对应同一个点，这样圆周上最终有三个点，这三个点构成等边三角形，该三角形的外接圆半径为\(\displaystyle 1\)，因此三角形中心（即原点）到三边的距离都是\(\displaystyle \frac{1}{2}\)．由题意知三角形有两个顶点的横坐标相同，即有一条边平行于\(\displaystyle y\)轴，这样三角形的三个顶点为
\[\displaystyle \left(\frac{1}{2},\frac{\sqrt{3}}{2}\right)，\left(\frac{1}{2},-\frac{\sqrt{3}}{2}\right)，\left(-1,0\right)\]或者
\[\displaystyle \left(-\frac{1}{2},\frac{\sqrt{3}}{2}\right)，\left(-\frac{1}{2},-\frac{\sqrt{3}}{2}\right)，\left(1,0\right)\]均有\(\displaystyle ab=-\frac{1}{2}\)．
思路3：不妨设\(\displaystyle \cos a_1=\cos a_2\)，于是\(\displaystyle x=\frac{a_1+a_2}{2}\)是曲线\(\displaystyle y=\cos x\)的对称轴，即\(\displaystyle \frac{a_1+a_2}{2}=k\pi\left(k\in\mathbb{Z}\right)\)，
代入\(\displaystyle a_2=a_1+\frac{2\pi}{3}\)，得\(\displaystyle a_1=-\frac{\pi}{3}+k\pi\left(k\in\mathbb{Z}\right)\)，\(\displaystyle a_2=\frac{\pi}{3}+k\pi\left(k\in\mathbb{Z}\right)\)，\(\displaystyle a_3=\pi+k\pi\left(k\in\mathbb{Z}\right)\)．
当\(\displaystyle k\)为奇数时，\(\displaystyle \cos a_1=\cos a_2=-\frac{1}{2}\)，\(\displaystyle \cos a_3=1\)，\(\displaystyle ab=-\frac{1}{2}\)；
当\(\displaystyle k\)为偶数时，\(\displaystyle \cos a_1=\cos a_2=\frac{1}{2}\)，\(\displaystyle \cos a_3=-1\)，\(\displaystyle ab=-\frac{1}{2}\)．
【实测数据】（河南）本题阅卷数据如下．
\[\displaystyle \begin{array}{c|cccccc} \text{题号} & \text{A} & \text{B} & \text{C} & \text{D} & \text{难度} & \text{区分度}\\ \hline \text{理10} & 10.63\% & 58.99\% & 15.79\% & 14.52\% & 0.59 & 0.42 \end{array}\]新答案（来源：1.28 等比数列.md）：D
【解题思路】当\(\displaystyle n\)为偶数且\(\displaystyle \left\{a_n\right\}\)的公比为\(\displaystyle -1\)时，\(\displaystyle X=Y=Z=0\)，此时四条式子都恒成立．
当\(\displaystyle n\)为奇数或\(\displaystyle \left\{a_n\right\}\)的公比不为\(\displaystyle -1\)时，\(\displaystyle X\)，\(\displaystyle Y-X\)，\(\displaystyle Z-Y\)成等比数列，则\(\displaystyle \left(Y-X\right)^2=X\left(Z-Y\right)\)，化简得\(\displaystyle Y^2-XY+X^2=XZ\)，即\(\displaystyle Y\left(Y-X\right)=X\left(Z-X\right)\)．
新答案（来源：1.29 数列综合.md）：C
【解题思路】由于\(\displaystyle \left\{a_n\right\}\)是公比为\(\displaystyle q\)的等比数列，
所以\(\displaystyle b_n=a_{m\left(n-1\right)+1}\left(1+q+\cdots+q^{m-1}\right)\)，
则\(\displaystyle \frac{b_{n+1}}{b_n}=\frac{a_{mn+1}\left(1+q+\cdots+q^{m-1}\right)}{a_{m\left(n-1\right)+1}\left(1+q+\cdots+q^{m-1}\right)}=\frac{a_{mn+1}}{a_{m\left(n-1\right)+1}}=q^m\)，
故\(\displaystyle \left\{b_n\right\}\)是公比为\(\displaystyle q^m\)的等比数列，\(\displaystyle A\)、\(\displaystyle B\)选项错误．
同理，\(\displaystyle c_n=a_{m\left(n-1\right)+1}^m\left(1\cdot q\cdot\cdots\cdot q^{m-1}\right)\)．
则\(\displaystyle \frac{c_{n+1}}{c_n}=\frac{a_{mn+1}^m\left(1\cdot q\cdot\cdots\cdot q^{m-1}\right)}{a_{m\left(n-1\right)+1}^m\left(1\cdot q\cdot\cdots\cdot q^{m-1}\right)}=\left(q^m\right)^m=q^{m^2}\)，
故\(\displaystyle \left\{c_n\right\}\)是公比为\(\displaystyle q^{m^2}\)的等比数列，故\(\displaystyle C\)选项正确，\(\displaystyle D\)选项错误．
新答案（来源：1.28 等比数列.md）：D
【解题思路】由矩形面积公式易得\(\displaystyle A_i=a_ia_{i+1}\)．先考虑\(\displaystyle \left\{A_n\right\}\)为等比数列的必要条件，若\(\displaystyle \left\{A_n\right\}\)是等比数列，则\(\displaystyle \frac{A_{n+1}}{A_n}=\frac{a_{n+1}a_{n+2}}{a_na_{n+1}}=q\)（\(\displaystyle q\)为非零常数），即\(\displaystyle \frac{a_{n+2}}{a_n}=q\)，当\(\displaystyle n\)为奇数时，得\(\displaystyle a_1,a_3,\cdots,a_{2n-1},\cdots\)是等比数列，公比为\(\displaystyle q\)；当\(\displaystyle n\)为偶数时，得\(\displaystyle a_2,a_4,\cdots,a_{2n},\cdots\)是等比数列，公比为\(\displaystyle q\)．因此，选项\(\displaystyle D\)是\(\displaystyle \left\{A_n\right\}\)为等比数列的必要条件．再考虑该必要条件是否充分，若\(\displaystyle \frac{a_{2n+1}}{a_{2n-1}}=q\)且\(\displaystyle \frac{a_{2n+2}}{a_{2n}}=q\)（\(\displaystyle q\)为非零常数），即\(\displaystyle \frac{a_{n+2}}{a_n}=q\)，又\(\displaystyle a_n\ne0\)，所以\(\displaystyle \frac{a_{n+1}a_{n+2}}{a_na_{n+1}}=q\)，即\(\displaystyle \frac{A_{n+1}}{A_n}=q\)，所以\(\displaystyle \left\{A_n\right\}\)为等比数列．因此，选项\(\displaystyle D\)又是\(\displaystyle \left\{A_n\right\}\)为等比数列的充分条件．
【实测数据】本题难度为\(\displaystyle 0.69\)，区分度为\(\displaystyle 0.32\)．各层次考生答题分布如下：
\[\displaystyle \begin{array}{c|cccccc} \text{考生层次} & \text{选A} & \text{选B} & \text{选C} & \text{选D} & \text{未选} & \text{得分率}\\ \hline \text{第一层次} & 2.00 & 0.53 & 8.74 & 88.73 & 0.00 & 0.89\\ \text{第二层次} & 6.01 & 1.93 & 17.66 & 74.36 & 0.04 & 0.75\\ \text{第三层次} & 10.33 & 3.35 & 22.91 & 63.30 & 0.11 & 0.64\\ \text{第四层次} & 12.57 & 7.52 & 30.89 & 48.80 & 0.22 & 0.49\\ \text{总体} & 7.67 & 3.27 & 19.90 & 69.07 & 0.09 & 0.69 \end{array}\]新答案（来源：1.26 数列的概念.md）：\(\displaystyle 4\)
【解题思路】令\(\displaystyle a_n=n\left(n+4\right)\left(\frac{2}{3}\right)^n\)，则
\[\displaystyle a_{n+1}-a_n=\left(n+1\right)\left(n+5\right)\left(\frac{2}{3}\right)^{n+1}-n\left(n+4\right)\left(\frac{2}{3}\right)^n=\left(\frac{2}{3}\right)^n\left[\frac{2}{3}\left(n+1\right)\left(n+5\right)-n\left(n+4\right)\right]=\left(\frac{2}{3}\right)^n\cdot\frac{10-n^2}{3}\]所以当\(\displaystyle n\leqslant3\)时，\(\displaystyle a_{n+1}-a_n>0\)，当\(\displaystyle n\geqslant4\)时，\(\displaystyle a_{n+1}-a_n<0\)，
从而\(\displaystyle a_1<a_2<a_3<a_4\)，且\(\displaystyle a_4>a_5>a_6>\cdots\)，故最大项是第\(\displaystyle 4\)项．
- 【2017上海15】已知\(\displaystyle a,b,c\)为实常数，数列\(\displaystyle \left \{ x_n \right \}\)的通项\(\displaystyle x_n=an^2+bn+c,n\in\mathbb{N^*}\)，则“存在\(\displaystyle k\in\mathbb{N^*}\)，使得\(\displaystyle x_{100+k},x_{200+k},x_{300+k}\)成等差数列”的一个必要条件是
8. 【2023全国乙卷10】已知等差数列\(\displaystyle \left \{ a_n \right \}\)的公差为\(\displaystyle d\)，集合\(\displaystyle S=\left \{\cos a_n\mid n\in\mathbb{N_+} \right \}\)。若\(\displaystyle d=\frac{2\pi}{3},S=\left \{ a,b \right \}\)，求\(\displaystyle ab\)；- \(\displaystyle a\geqslant 0\)
- \(\displaystyle b\leqslant 0\)
- \(\displaystyle c=0\)
- \(\displaystyle a-2b+c=0\)
答案
\(\displaystyle -1/2.\) 将等差数列的项\(\displaystyle a_n\)对应为单位圆周上的点\(\displaystyle (\cos a_n, \sin a_n)\)，由于公差是\(\displaystyle \frac{2\pi}{3}\)，即\(\displaystyle a_{n+3k}\)与\(\displaystyle a_n\)对应同一个点，这样圆周上最终有三个点，这三个点构成等边三角形，该三角形的外接圆半径为 1，因此三角形中心（即原点）到三边的距离都是\(\displaystyle \frac{1}{2}\). 由题意知三角形有两个顶点的横坐标相同，即有一条边平行于\(\displaystyle y\)轴，这样三角形的三个顶点为\(\displaystyle \left\{\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right), \left(\frac{1}{2}, -\frac{\sqrt{3}}{2}\right), (-1, 0)\right\}\)或者\(\displaystyle \left\{\left(-\frac{1}{2}, \frac{\sqrt{3}}{2}\right), \left(-\frac{1}{2}, -\frac{\sqrt{3}}{2}\right), (1, 0)\right\}\)，均有\(\displaystyle ab=-\frac{1}{2}\).
- 【2010安徽理10】设 \(\displaystyle \{a_n\}\) 是任意等比数列, 其前 \(\displaystyle n\) 项和为\(\displaystyle S_n\),对任意正整数\(\displaystyle n\)，下列命题中正确的是
10. 【2021“数海漫游一模”（网络联考）10】已知实数\(\displaystyle x,y,z\)满足\(\displaystyle 0<x<y<z<\pi\)，且\(\displaystyle \sin x,\sin y,\sin z\)可以按照某个顺序排成等差数列，则- \(\displaystyle S_n+S_{3n}=S_{2n}\)
- \(\displaystyle S_{2n}(S_{2n}-S_n)=S_{3n}(S_{3n}-S_{n})\)
- \(\displaystyle S^2_{2n}=S_{n}S_{3n}\)
- \(\displaystyle S_{2n}(S_{2n}-S_n)=S_n(S_{3n}-S_n)\)
11. 已知数列\(\displaystyle \left \{ a_n \right \}\)和\(\displaystyle \left \{ b_n \right \}\)的通项公式为\(\displaystyle a_n=17n-4,b_n=-2\cdot (-\frac{3}{4})^n\)，问数列\(\displaystyle \left \{ a_nb_n\right \}\)是否有最大项和最小项。 12. 【2025成都一诊8】等比数列\(\displaystyle \left \{ a_n \right \}\)中，\(\displaystyle a_1+a_2+a_3+a_4=1,|a_1|+|a_2|+|a_3|+|a_4|=3\)，求\(\displaystyle a_1\)。 13. 【2007湖北理8】 已知等差数列 \(\displaystyle \{a_n\},\{b_n\}\) 的前 \(\displaystyle n\) 项和分别为 \(\displaystyle A_n,B_n\)，且\(\displaystyle \frac{A_n}{B_n} = \frac{7n+45}{n+3}\), 求所有使得 \(\displaystyle \frac{a_n}{b_n}\in \mathbb{Z}\)的正整数 \(\displaystyle n\)。 14. 【2013福建理9】已知等比数列 \(\displaystyle \{a_n\}\) 的公比为 \(\displaystyle q\), 记 \(\displaystyle b_n=a_{m(n-1)+1}+a_{m(n-1)+2}+\cdots+a_{m(n-1)+m}\), \(\displaystyle c_n=a_{m(n-1)+1} \cdot a_{m(n-1)+2} \cdot \cdots \cdot a_{m(n-1)+m}\), (\(\displaystyle m,n \in \mathbb{N}^*\)), 则下述命题中正确的是- \(\displaystyle \sin x\)不可能是等差中项
- \(\displaystyle \sin y\)不可能是等差中项
- \(\displaystyle \sin z\)不可能是等差中项
- \(\displaystyle \sin x,\sin y,\sin z\)都可能是等差中项
15. 【2004湖北理8】设数列 \(\displaystyle \{a_n\}\) 的前 \(\displaystyle n\) 项和 \(\displaystyle S_n = a\left[2 - \left(\frac{1}{2}\right)^{n-1}\right] - b\left[2 - (n+1)\left(\frac{1}{2}\right)^{n-1}\right]\) (\(\displaystyle n=1,2,\cdots\)), 其中 \(\displaystyle a,b\) 是非零常数. 则存在数列 \(\displaystyle \{x_n\},\{y_n\}\) 使得- 数列 \(\displaystyle \{b_n\}\) 为等差数列, 公差为 \(\displaystyle q^m\)
- 数列 \(\displaystyle \{b_n\}\) 为等比数列, 公比为 \(\displaystyle q^{2m}\)
- 数列 \(\displaystyle \{c_n\}\) 为等比数列, 公比为 \(\displaystyle q^{m^2}\)
- 数列 \(\displaystyle \{c_n\}\) 为等比数列, 公比为 \(\displaystyle q^{m^m}\)
- \(\displaystyle a_n = x_n + y_n\), 其中 \(\displaystyle \{x_n\}\) 为等差数列, \(\displaystyle \{y_n\}\) 为等比数列
- \(\displaystyle a_n = x_n + y_n\), 其中 \(\displaystyle \{x_n\}\) 和 \(\displaystyle \{y_n\}\) 都为等差数列
- \(\displaystyle a_n = x_n \cdot y_n\), 其中 \(\displaystyle \{x_n\}\) 为等差数列, \(\displaystyle \{y_n\}\) 为等比数列
- \(\displaystyle a_n = x_n \cdot y_n\), 其中 \(\displaystyle \{x_n\}\) 和 \(\displaystyle \{y_n\}\) 都为等比数列
答案
新答案（来源：3.2 性质的证明（2）数列.md）： 必要性．设\(\displaystyle \left\{a_n\right\}\)是公差为\(\displaystyle d_1\)的等差数列，则 $\(\displaystyle b_{n+1}-b_n=\left(a_{n+1}-a_{n+3}\right)-\left(a_n-a_{n+2}\right)=\left(a_{n+1}-a_n\right)-\left(a_{n+3}-a_{n+2}\right)=d_1-d_1=0,\)$ 所以\(\displaystyle b_n\leqslant b_{n+1}\left(n=1,2,3,\cdots\right)\)成立．
又 $\(\displaystyle c_{n+1}-c_n=\left(a_{n+1}-a_n\right)+2\left(a_{n+2}-a_{n+1}\right)+3\left(a_{n+3}-a_{n+2}\right)\)$ $\(\displaystyle =d_1+2d_1+3d_1=6d_1\text{（常数）}\left(n=1,2,3,\cdots\right),\)$ 所以数列\(\displaystyle \left\{c_n\right\}\)为等差数列．
充分性．设数列\(\displaystyle \left\{c_n\right\}\)是公差为\(\displaystyle d_2\)的等差数列，且\(\displaystyle b_n\leqslant b_{n+1}\left(n=1,2,3,\cdots\right)\)．
证法一：因为\(\displaystyle c_n=a_n+2a_{n+1}+3a_{n+2}\)，
①
所以\(\displaystyle c_{n+2}=a_{n+2}+2a_{n+3}+3a_{n+4}\)．
②
①\(\displaystyle -\)②得\(\displaystyle c_n-c_{n+2}=\left(a_n-a_{n+2}\right)+2\left(a_{n+1}-a_{n+3}\right)+3\left(a_{n+2}-a_{n+4}\right)=b_n+2b_{n+1}+3b_{n+2}\)．
因为\(\displaystyle c_n-c_{n+2}=\left(c_n-c_{n+1}\right)+\left(c_{n+1}-c_{n+2}\right)=-2d_2\)，
所以\(\displaystyle b_n+2b_{n+1}+3b_{n+2}=-2d_2\)，
③
从而有\(\displaystyle b_{n+1}+2b_{n+2}+3b_{n+3}=-2d_2\)．
④
④\(\displaystyle -\)③得\(\displaystyle \left(b_{n+1}-b_n\right)+2\left(b_{n+2}-b_{n+1}\right)+3\left(b_{n+3}-b_{n+2}\right)=0\)．
⑤
因为\(\displaystyle b_{n+1}-b_n\geqslant0\)，\(\displaystyle b_{n+2}-b_{n+1}\geqslant0\)，\(\displaystyle b_{n+3}-b_{n+2}\geqslant0\)，
所以由⑤得\(\displaystyle b_{n+1}-b_n=0\left(n=1,2,3,\cdots\right)\)．
由此不妨设\(\displaystyle b_n=d_3\left(n=1,2,3,\cdots\right)\)，则\(\displaystyle a_n-a_{n+2}=d_3\)（常数）．
由此\(\displaystyle c_n=a_n+2a_{n+1}+3a_{n+2}=4a_n+2a_{n+1}-3d_3\)，
从而\(\displaystyle c_{n+1}=4a_{n+1}+2a_{n+2}-3d_3=4a_{n+1}+2a_n-5d_3\)．
两式相减得\(\displaystyle c_{n+1}-c_n=2\left(a_{n+1}-a_n\right)-2d_3\)，
因此\(\displaystyle a_{n+1}-a_n=\frac{1}{2}\left(c_{n+1}-c_n\right)+d_3=\frac{1}{2}d_2+d_3\)（常数）\(\displaystyle \left(n=1,2,3,\cdots\right)\)，
所以数列\(\displaystyle \left\{a_n\right\}\)是等差数列．
证法二：令\(\displaystyle A_n=a_{n+1}-a_n\)，由\(\displaystyle b_n\leqslant b_{n+1}\)知\(\displaystyle a_n-a_{n+2}\leqslant a_{n+1}-a_{n+3}\)，
从而\(\displaystyle a_{n+1}-a_n\geqslant a_{n+3}-a_{n+2}\)，即\(\displaystyle A_n\geqslant A_{n+2}\left(n=1,2,3,\cdots\right)\)．
由\(\displaystyle c_n=a_n+2a_{n+1}+3a_{n+2}\)，\(\displaystyle c_{n+1}=a_{n+1}+2a_{n+2}+3a_{n+3}\)得 $\(\displaystyle c_{n+1}-c_n=\left(a_{n+1}-a_n\right)+2\left(a_{n+2}-a_{n+1}\right)+3\left(a_{n+3}-a_{n+2}\right)，\)$ 即 $\(\displaystyle A_n+2A_{n+1}+3A_{n+2}=d_2.\)$ ⑥ 由此得\(\displaystyle A_{n+2}+2A_{n+3}+3A_{n+4}=d_2\)．
⑦
⑥\(\displaystyle -\)⑦得\(\displaystyle \left(A_n-A_{n+2}\right)+2\left(A_{n+1}-A_{n+3}\right)+3\left(A_{n+2}-A_{n+4}\right)=0\)．
⑧
因为\(\displaystyle A_n-A_{n+2}\geqslant0\)，\(\displaystyle A_{n+1}-A_{n+3}\geqslant0\)，\(\displaystyle A_{n+2}-A_{n+4}\geqslant0\)，
所以由⑧得\(\displaystyle A_n-A_{n+2}=0\left(n=1,2,3,\cdots\right)\)．
于是由⑥得\(\displaystyle 4A_n+2A_{n+1}=A_n+2A_{n+1}+3A_{n+2}=d_2\)，
⑨
从而\(\displaystyle 2A_n+4A_{n+1}=4A_{n+1}+2A_{n+2}=d_2\)．
⑩
由⑨和⑩得\(\displaystyle 4A_n+2A_{n+1}=2A_n+4A_{n+1}\)，故\(\displaystyle A_n=A_{n+1}\)，
即\(\displaystyle a_{n+2}-a_{n+1}=a_{n+1}-a_n\left(n=1,2,3,\cdots\right)\)．
所以数列\(\displaystyle \left\{a_n\right\}\)是等差数列．
- 【2002上海16】 设等差数列 \(\displaystyle \{a_n\}\)的前\(\displaystyle n\)项和为 \(\displaystyle S_n\), 且 \(\displaystyle S_5 < S_6\), \(\displaystyle S_6 = S_7 > S_8\), 则下列结论错误的是
17. 【2011上海理18】设 \(\displaystyle \{a_n\}\) 是正项无穷数列, \(\displaystyle A_i=a_ia_{i+1}\)(\(\displaystyle i=1,2,\cdots\)), 则 \(\displaystyle \{A_n\}\) 为等比数列的充要条件是- \(\displaystyle d < 0\)
- \(\displaystyle a_7 = 0\)
- \(\displaystyle S_9 > S_5\)
- \(\displaystyle S_6\) 和 \(\displaystyle S_7\) 均为 \(\displaystyle S_n\) 的最大值
18. （多选）设\(\displaystyle S_n\)是等比数列\(\displaystyle \left \{ a_n\right \}\)的前\(\displaystyle n\)项和，\(\displaystyle q\)为\(\displaystyle \left \{ a_n\right \}\)的公比，则- \(\displaystyle \{a_n\}\) 是等比数列
- \(\displaystyle a_1,a_3,\cdots,a_{2n-1},\cdots\) 或 \(\displaystyle a_2,a_4,\cdots,a_{2n},\cdots\) 是等比数列
- \(\displaystyle a_1,a_3,\cdots,a_{2n-1},\cdots\) 和 \(\displaystyle a_2,a_4,\cdots,a_{2n},\cdots\) 均是等比数列
- \(\displaystyle a_1,a_3,\cdots,a_{2n-1},\cdots\) 和 \(\displaystyle a_2,a_4,\cdots,a_{2n},\cdots\) 均是等比数列, 且公比相同
19. 【2025“漫游数海”回归课本联赛10】（多选）已知\(\displaystyle a,b,c\)均为正整数，关于\(\displaystyle x\)的一元二次方程\(\displaystyle (a-b)x^2+(a-c)x+(b-c)=0\)仅有一个实根\(\displaystyle x=x_0\)，则- \(\displaystyle \left \{ a_n^2\right \}\)为等比数列
- \(\displaystyle \left \{ qS_n\right \}\)为等比数列
- 若\(\displaystyle q=-1\)，则存在\(\displaystyle m\in\mathbb{N^*}\)使得\(\displaystyle S_m=0\)
- 若存在\(\displaystyle m\in\mathbb{N^*}\)使得\(\displaystyle S_m=0\)，则\(\displaystyle q=-1\)
20. 【2004上海文12】（多选） 若干个能唯一确定一个数列的量称为该数列的“基本量”. 设 \(\displaystyle \{a_n\}\) 是公比为 \(\displaystyle q\) 的无穷等比数列,\(\displaystyle S_n\) 为 \(\displaystyle \{a_n\}\) 的前 \(\displaystyle n\) 项和。 下列 \(\displaystyle \{a_n\}\) 的四组量中, 一定能成为该数列“基本量”的是- \(\displaystyle a,b,c\)成等差数列
- \(\displaystyle a,b,c\)成等比数列
- \(\displaystyle x_0^a,x_0^b,x_0^c\)成等比数列
- \(\displaystyle a^{x_0},b^{x_0},c^{x_0}\)成等差数列
21. 【2016上海11】无穷数列\(\displaystyle \left \{ a_n \right \}\)由\(\displaystyle k\)个不同的数组成，\(\displaystyle S_n\)为\(\displaystyle \left \{ a_n \right \}\)的前\(\displaystyle n\)项和，若对任意正整数\(\displaystyle n\),\(\displaystyle S_n\in\left \{ 2,3 \right \}\)，求\(\displaystyle k\)的最大值。 22. 【2021 新高考I卷16】规格为 \(\displaystyle 20 \times 12\)的长方形纸对折 \(\displaystyle 1\) 次一共可以得到 \(\displaystyle 10\times 12 , 20 \times 6\) 这两种规格的图形，它们的面积之和 \(\displaystyle S_1 = 240\)，对折 \(\displaystyle 2\) 次共可以得到 \(\displaystyle 5 \times 12, 10 \times 6, 20 \times 3\) 这三种规格的图形，它们的面积之和 \(\displaystyle S_2 = 180\cdots\)求\(\displaystyle \sum_{k=1}^n S_k\)。 23. 【2011浙江文17】若数列 \(\displaystyle \left\{n(n+4)\left(\frac{2}{3}\right)^n\right\}\) 中的最大项是第 \(\displaystyle k\) 项, 求\(\displaystyle k\)。- \(\displaystyle S_1\) 与 \(\displaystyle S_2\)
- \(\displaystyle a_2\) 与 \(\displaystyle S_3\)
- \(\displaystyle a_1\) 与 \(\displaystyle a_n\)
- \(\displaystyle q\) 与 \(\displaystyle a_n\).
答案
\(\displaystyle 4\).
令$\displaystyle a_n=n\left(n+4\right)\left(\frac23\right)^n$，则\[\displaystyle \begin{aligned} a_{n+1}-a_n &=\left(n+1\right)\left(n+5\right)\left(\frac23\right)^{n+1}-n\left(n+4\right)\left(\frac23\right)^n\\ &=\left(\frac23\right)^n\left[\frac23\left(n+1\right)\left(n+5\right)-n\left(n+4\right)\right]\\ &=\left(\frac23\right)^n\cdot\frac{10-n^2}{3}. \end{aligned}\]所以当\(\displaystyle n\leqslant3\)时，\(\displaystyle a_{n+1}-a_n>0\)，当\(\displaystyle n\geqslant4\)时，\(\displaystyle a_{n+1}-a_n<0\)，
从而\(\displaystyle a_1<a_2<a_3<a_4\)，且\(\displaystyle a_4>a_5>a_6>\cdots\)，故最大项是第 4 项。
- 互不相等的正数\(\displaystyle a,b,c\)构成等比数列，若\(\displaystyle \log_ab,\log_bc,\log_ca\)成等差数列，求其公差。
- 对下列题设，判断命题\(\displaystyle p\)是命题\(\displaystyle q\)的充分不必要条件、必要不充分条件、充要条件，还是既不充分也不必要条件。
- 已知\(\displaystyle \left \{ a_n\right \}\)是公比为\(\displaystyle q\)的等比数列。命题\(\displaystyle p\): \(\displaystyle q>0\)；命题\(\displaystyle q\): \(\displaystyle \left \{ a_n\right \}\)递增。
- 已知数列 \(\displaystyle \{a_n\}\) 是等比数列：命题\(\displaystyle p\)： 对任意正整数\(\displaystyle n\)，\(\displaystyle a_{n+2} > a_n\)；命题\(\displaystyle q\)：\(\displaystyle \{a_n\}\) 递增。
- 设\(\displaystyle a_1,a_2,\cdots ,a_n\in\mathbb{R}，n\geqslant 3\)：命题\(\displaystyle p\): \(\displaystyle a_1,a_2,\cdots a_n\)成等比数列；命题\(\displaystyle q\)：\(\displaystyle (a_1^2+a^2_2+\cdots a_{n-1}^2)(a_2^2+a_3^2+\cdots a_n^2)=(a_1a_2+a_2a_3+\cdots a_{n-1}a_n)^2\)
- 已知无穷等比数列\(\displaystyle \{a_n\}\)的前\(\displaystyle n\)项和为\(\displaystyle S_n\)： 命题\(\displaystyle p\): \(\displaystyle a_1a_2>a_2^2\)；命题\(\displaystyle q\): \(\displaystyle S_n\)既无最大值也无最小值
- 已知\(\displaystyle \left \{ a_n \right \}\)为等比数列： 命题\(\displaystyle p\): \(\displaystyle a_{2024}=1\)；命题\(\displaystyle q\): 对任意正整数\(\displaystyle n\)，\(\displaystyle a_1a_2\cdots a_n=a_1a_2\cdots a_{4047-n}\)
- 已知\(\displaystyle \left \{ a_n\right \}\)是无穷数列 命题\(\displaystyle p\): \(\displaystyle \left \{ a_n \right \}\)存在最小项；命题\(\displaystyle q\): \(\displaystyle \left \{ a_n+a_{n+1} \right \}\)存在最小项
- 已知\(\displaystyle \left \{ a_n \right \}\)是等差数列： 命题\(\displaystyle p\): \(\displaystyle a_5,a_7,a_8\)构成公比不为1的等比数列；命题\(\displaystyle q\): \(\displaystyle a_9=0\)
- 已知正整数\(\displaystyle i,j\)，记\(\displaystyle S_n\)为等比数列\(\displaystyle \left \{ a_n \right \}\)的前\(\displaystyle n\)项和： 命题\(\displaystyle p\): \(\displaystyle a_{2i+1}<0\)；命题\(\displaystyle q\): \(\displaystyle S_{2j+1}<0\)
- 已知等差数列\(\displaystyle \left \{ a_n \right \}\)的各项均不为0，记\(\displaystyle S_n\)为\(\displaystyle \left \{a_n \right \}\)的前\(\displaystyle n\)项和： 命题\(\displaystyle p\)：“\(\displaystyle a_4a_6<0\)”；命题\(\displaystyle q\)：对任意正整数\(\displaystyle n\)，\(\displaystyle \frac{S_5}{a_5}\leqslant \frac{S_n}{a_n}\)
- 已知数列\(\displaystyle \left \{ a_n \right \}\)满足 $\(\displaystyle a_{n+1}= \begin{cases} 3a_n-1,a_n\text{为奇数} \\ \frac{a_n}{2},a_n\text{为偶数} \end{cases}\)$
- 若\(\displaystyle a_1=6\)，求\(\displaystyle a_1+a_2+\cdots a_{2024}\)；
- 若\(\displaystyle a_6=5\)，求\(\displaystyle a_1\)的所有可能的取值；
- 若\(\displaystyle a_1\in\mathbb{N^*}\)，是否一定存在\(\displaystyle m\in\mathbb{N^*}\)，使得\(\displaystyle a_m=1\)？试说明理由。
- 等差数列\(\displaystyle \left \{ a_n \right \}\)与等比数列\(\displaystyle \left \{ b_n \right \}\)均为无穷正项实数列，且满足\(\displaystyle a_1=b_1,a_2=b_2\)，证明： $\(\displaystyle (a).a_{n+1}\geqslant a_n\quad (b).a_n\leqslant b_n\)$
-
【2005湖南理20】自然状态下的鱼类是一种可再生的资源，为持续利用这一资源，需从宏观上考察其再生能力及捕捞强度对鱼群总量的影响。用\(\displaystyle x_n\)表示某鱼群在第\(\displaystyle n\)年年初时的总量，\(\displaystyle n\in\mathbb{N^*}\)，且\(\displaystyle x_1>0\)，不考虑其他因素，设在第\(\displaystyle n\)年内鱼群的繁殖量及被捕捞量都与\(\displaystyle x_n\)成正比，死亡量与\(\displaystyle x_n^2\)成正比，这些比例系数依次为正常数\(\displaystyle a,b,c\)。
-
求\(\displaystyle x_n\)与\(\displaystyle x_{n+1}\)的递推关系式；
- 猜测：当且仅当\(\displaystyle x_1,a,b,c\)满足什么条件时，每年年初鱼群的总量保持不变（不要求证明）；
- 设\(\displaystyle a=2,c=1\)，为保证对一切\(\displaystyle x_1\in(0,2)\)的值，都有\(\displaystyle x_n>0,n\in\mathbb{N^*}\)，求捕捞强度\(\displaystyle b\)的最大允许值。
- 【2007安徽文21】某国采用养老储备金制度. 公民在就业的第一年就交纳养老储备金, 数目为 \(\displaystyle a_1\), 以后每年交纳的数目均比上一年增加 \(\displaystyle d\) (\(\displaystyle d>0\)), 因此, 历年所交纳的储备金数目 \(\displaystyle a_1,a_2,\cdots\) 是一个公差为 \(\displaystyle d\) 的等差数列, 与此同时, 国家给予优惠的计息政策, 不仅采用固定利率, 而且计算复利. 这就是说, 如果固定年利率为 \(\displaystyle r\) (\(\displaystyle r>0\)), 那么, 在第 \(\displaystyle n\) 年末, 第一年所交纳的储备金就变为 \(\displaystyle a_1(1+r)^{n-1}\), 第二年所交纳的储备金就变为 \(\displaystyle a_2(1+r)^{n-2}\), \(\displaystyle \cdots\). 以 \(\displaystyle T_n\) 表示到第 \(\displaystyle n\) 年末所累计的储备金总额。
- 写出 \(\displaystyle T_n\) 与 \(\displaystyle T_{n-1}\) (\(\displaystyle n \geqslant 2\)) 的递推关系式；
- 求证: \(\displaystyle T_n = A_n + B_n\), 其中 \(\displaystyle \{A_n\}\) 是一个等比数列, \(\displaystyle \{B_n\}\) 是一个等差数列。
- 【2014湖南卷理20】已知数列\(\displaystyle \left \{ a_n \right \}\)满足\(\displaystyle a_1=1,|a_{n+1}-a_n|=p,n\in\mathbb{N^*}\)。
- 若\(\displaystyle \left \{ a_n \right \}\)是递增数列，\(\displaystyle a_1,2a_2,3a_3\)成等差数列，求\(\displaystyle p\)；
- 若\(\displaystyle p=\frac{1}{2}\)，且\(\displaystyle \left \{ a_{2n-1} \right \}\)是递增数列，\(\displaystyle \left \{ a_{2n} \right \}\)是递减数列，\(\displaystyle a_2>a_1\)，求\(\displaystyle \left \{ a_n \right \}\)的通项公式。
- 【1995全国卷理25】设\(\displaystyle \{a_n\}\)是由正数组成的等比数列，\(\displaystyle S_n\)是其前\(\displaystyle n\)项和。是否存在常数\(\displaystyle c>0\)，使得\(\displaystyle \frac{\lg(S_n-c)+\lg(S_{n+2}-c)}{2}=\lg(S_{n+1}-c)\)对任意正整数\(\displaystyle n\)成立？
- 【2004江苏20】设无穷等差数列\(\displaystyle \left \{ a_n \right \}\)的前\(\displaystyle n\)项和为\(\displaystyle S_n\)。求所有的\(\displaystyle \left \{ a_n \right \}\)，使得对于一切正整数\(\displaystyle k\)都有\(\displaystyle S_{k^2}=(S_k)^2\)成立。
- 【2001京蒙皖理20】在 1 与 2 之间插入 \(\displaystyle n\) 个正数 \(\displaystyle a_1,a_2,a_3,\cdots,a_n\), 使这 \(\displaystyle n+2\) 个数成等比数列； 又在 1 与 2 之间插入 \(\displaystyle n\) 个正数 \(\displaystyle b_1,b_2,b_3,\cdots,b_n\), 使这 \(\displaystyle n+2\) 个数成等差数列. 记 \(\displaystyle A_n = a_1a_2a_3\cdots a_n\), \(\displaystyle B_n = b_1 + b_2 + b_3 + \cdots + b_n\)。当 \(\displaystyle n \geqslant 7\) 时, 比较 \(\displaystyle A_n\) 与 \(\displaystyle B_n\) 的大小。
- 【2023新高考I卷20】设等差数列\(\displaystyle \left \{ a_n \right \}\)的公差为\(\displaystyle d\)，\(\displaystyle d>1\)。令\(\displaystyle b_n=\frac{n^2+n}{a_n}\)，记\(\displaystyle S_n,T_n\)为别为数列\(\displaystyle \left \{ a_n \right \} ,\left \{ b_n \right \}\)的前\(\displaystyle n\)项和。若\(\displaystyle \left \{ b_n \right \}\)为等差数列，且\(\displaystyle S_{99}-T_{99}=99\)，求\(\displaystyle d\)。
- 设数列 \(\displaystyle \{a_n\}\) 各项均为实数，且当 \(\displaystyle n \geqslant 2\) 时，\(\displaystyle a_{n+1} = \lvert a_n \rvert - a_{n-1}\)。
已知：若存在 \(\displaystyle k \in \mathbb{N}^*\) 使得 \(\displaystyle a_k = a_{k+T}\) 且 \(\displaystyle a_{k+1} = a_{k+1+T}\) （\(\displaystyle T \in \mathbb{N}^*\)），则 \(\displaystyle \{a_n\}\) 是周期为 \(\displaystyle T\) 的数列。（注：这个结论其实可用归纳法两行证完） 证明：
（1） 存在大于 1 的正整数 \(\displaystyle m\)，使得 \(\displaystyle a_m \leqslant 0\)；
（2） 存在正整数 \(\displaystyle m\)，使得 \(\displaystyle a_m \leqslant 0\) 且 \(\displaystyle a_{m+1} \leqslant 0\)；
（3） 对任意正整数 \(\displaystyle n\)，都有 \(\displaystyle a_{n+9} = a_n\)。 36. 【2006江苏21】 设数列 \(\displaystyle \{a_n\},\{b_n\},\{c_n\}\) 满足：对任意正整数\(\displaystyle n\)，\(\displaystyle b_n = a_n - a_{n+2}\), \(\displaystyle c_n = a_n + 2a_{n+1} + 3a_{n+2}\)，证明: \(\displaystyle \{a_n\}\) 为等差数列的充要条件是： \(\displaystyle \{c_n\}\) 为等差数列且 对任意正整数\(\displaystyle n\)，\(\displaystyle b_n \leqslant b_{n+1}\)。 37. 【2005全国III卷理20】 在等差数列 \(\displaystyle \{a_n\}\) 中, 公差 \(\displaystyle d \neq 0\), \(\displaystyle a_2\) 是 \(\displaystyle a_1\) 与 \(\displaystyle a_4\) 的等比中项. 已知数列 \(\displaystyle a_1,a_3,a_{k_1},a_{k_2},\dots,a_{k_n},\dots\) 成等比数列, 求数列 \(\displaystyle \{k_n\}\) 的通项 \(\displaystyle k_n\)。 38. 【2024广东二模18】已知数列 \(\displaystyle \{a_n\}\)满足\(\displaystyle a_1 = 1\)，且对任意正整数\(\displaystyle k\)，\(\displaystyle a_{2k-1}, a_{2k}, a_{2k+1}\) 成公差为 \(\displaystyle 2k\)的等差数列。是否存在 \(\displaystyle x\)，使得对任意正整数\(\displaystyle k\)，\(\displaystyle a_{2k} + x, a_{2k+1} + x, a_{2k+2} + x\) 成等比数列？ 39. 【2001上海22】已知 \(\displaystyle \{a_n\}\) 是首项为 2, 公比为 \(\displaystyle \frac{1}{2}\) 的等比数列, \(\displaystyle S_n\) 为它的前 \(\displaystyle n\) 项和。是否存在自然数 \(\displaystyle c,k\)，使得 \(\displaystyle \frac{S_{k+1} - c}{S_k - c} > 2\) 恒成立？
B 组习题
习题组II
-
【2024北京15改编】（多选）设\(\displaystyle \left \{ a_n \right \} ,\left \{ b_n \right \}\)是不同的无穷数列，且都不是常数列，记集合\(\displaystyle M=\left \{ k\mid a_k=b_k,k\in\mathbb{N^*} \right \}\)，下列命题中正确的是
- 若\(\displaystyle \left \{ a_n \right \}\)与\(\displaystyle \left \{ b_n \right \}\)均为等差数列，则\(\displaystyle M\)中最多有1个元素
- 若\(\displaystyle \left \{ a_n \right \}\)与\(\displaystyle \left \{ b_n \right \}\)均为等比数列，则\(\displaystyle M\)中最多有2个元素
- 若\(\displaystyle \left \{ a_n \right \}\)为等差数列，\(\displaystyle \left \{ b_n \right \}\)为等比数列，则\(\displaystyle M\)中最多有3个元素
- 若\(\displaystyle \left \{ a_n \right \}\)为递增数列，\(\displaystyle \left \{ b_n \right \}\)为递减数列，则\(\displaystyle M\)中最多有1个元素
答案
ACD. 设\(\displaystyle a_n=a+nd\)，\(\displaystyle b_n=b+nc\). 则\(\displaystyle k\in M\)当且仅当\(\displaystyle a+kd=b+kc\)，即\(\displaystyle a-b=k\left(c-d\right)\). 这是个一次方程，这个关于\(\displaystyle k\)的方程只可能有\(\displaystyle 0\)个或\(\displaystyle 1\)个正整数解. 特别地，对\(\displaystyle a_n=2n-1\)，\(\displaystyle b_n=3n-2\)，此时\(\displaystyle M=\left\{1\right\}\). 故①正确.
②考虑\(\displaystyle a_n=2^n\)，\(\displaystyle b_n=\left(-2\right)^n\)，则\(\displaystyle \left\{a_n\right\}\)与\(\displaystyle \left\{b_n\right\}\)都是等比数列，但是\(\displaystyle M=\left\{2,4,6,8,\cdots\right\}=\left\{\text{全体偶数}\right\}\)，故②错误.
③设\(\displaystyle a_n=a+nd\)，\(\displaystyle b_n=q^n\)（不妨设\(\displaystyle b_1=q\)）. 则\(\displaystyle k\in M\)当且仅当\(\displaystyle a+kd=q^k\).
设\(\displaystyle f\left(x\right)=\left|q\right|^x-dx-a\)，\(\displaystyle g\left(x\right)=\left|q\right|^x+dx+a\).
（i）若\(\displaystyle q>0\)，则\(\displaystyle a+kd=q^k\)当且仅当\(\displaystyle f\left(k\right)=0\). 求导可知\(\displaystyle f\left(x\right)\)最多只有两段单调区间，从而最多只有\(\displaystyle 2\)个零点，即\(\displaystyle M\)的元素个数不超过\(\displaystyle 2\).
（ii）若\(\displaystyle q<0\)，则当\(\displaystyle k\)为奇数时，\(\displaystyle a+kd=q^k\)等价于\(\displaystyle a+kd=-\left|q\right|^k\)，等价于\(\displaystyle g\left(k\right)=0\).
当\(\displaystyle k\)为偶数时，\(\displaystyle a+kd=q^k\)等价于\(\displaystyle f\left(k\right)=0\).
注意到，通过讨论\(\displaystyle q,d\)的取值可发现，\(\displaystyle f\left(x\right)\)和\(\displaystyle g\left(x\right)\)必有一个在\(\displaystyle \mathbb{R}\)上为单调函数，另一个在\(\displaystyle \mathbb{R}\)上有两段单调区间. 于是，\(\displaystyle f\left(x\right),g\left(x\right)\)的零点个数之和不超过\(\displaystyle 3\).
特别地，取\(\displaystyle b_n=\left(-\frac12\right)^n\)，则\(\displaystyle b_1=-\frac12\)，\(\displaystyle b_3=-\frac18\)，\(\displaystyle b_4=\frac1{16}\). 再取\(\displaystyle a_n=-\frac12+\frac3{16}\left(n-1\right)\)，则\(\displaystyle a_1=-\frac12\)，\(\displaystyle a_3=-\frac18\)，\(\displaystyle a_4=\frac1{16}\)，所以\(\displaystyle M=\left\{1,3,4\right\}\). 所以③正确.
④设\(\displaystyle c_n=a_n-b_n\)，则\(\displaystyle \left\{c_n\right\}\)是递增数列，\(\displaystyle M=\left\{n\mid c_n=0\right\}\). 取一个连续函数\(\displaystyle f\left(x\right)\)满足\(\displaystyle f\left(n\right)=c_n\)，使得\(\displaystyle f\left(x\right)\)是单调递增函数（构造方法：连接\(\displaystyle \left(n,c_n\right)\)和\(\displaystyle \left(n+1,c_{n+1}\right)\)的折线）. 由零点存在定理，\(\displaystyle f\left(x\right)\)最多只有一个零点，\(\displaystyle M\)最多只有\(\displaystyle 1\)个元素. 故④正确.
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答案
\(\displaystyle \sqrt[3]{\frac{3}{2}}.\) i.对于任意 \(\displaystyle n\in\mathbb{N}^{*}\)，均有 \(\displaystyle a_1+a_2+\cdots+a_{3n}=n^2+n\) 等价于相邻三项之和是给定的：\(\displaystyle a_{3n-2}+a_{3n-1}+a_{3n}=\left(n+1\right)n-n\left(n-1\right)=2n\)。 ii.\(\displaystyle \left\{a_n\right\}\) 中有某连续 \(\displaystyle 9\) 项 \(\displaystyle a_k,a_{k+1},\cdots,a_{k+8}\) 是公比为 \(\displaystyle q\) 的等比数列 意味着必有连续的 \(\displaystyle 6\) 项 \(\displaystyle a_{3m-2},a_{3m-1},a_{3m},a_{3m+1},a_{3m+2},a_{3m+3}\) 是公比为 \(\displaystyle q\) 的等比数列。 其中 \(\displaystyle a_{3m-2}+a_{3m-1}+a_{3m}=2m\)， \(\displaystyle a_{3m+1}+a_{3m+2}+a_{3m+3}=2m+2\)，
所以 \(\displaystyle \left(2m+2\right)=q^3\cdot2m\)，即 \(\displaystyle \left(q^3-1\right)m=1\)。 又 \(\displaystyle m\) 是正整数，所以 \(\displaystyle q^3-1=\frac1m\)，\(\displaystyle q=\sqrt[3]{\frac{m+1}{m}}\)。
接下来从大到小讨论 \(\displaystyle q\) 的所有可能取值。
（1）如果 \(\displaystyle m=1\)，则 \(\displaystyle q=\sqrt[3]{2}\)，我们尝试构造满足①②的数列 \(\displaystyle \left\{a_n\right\}\)，
所以 \(\displaystyle a_1+a_2+a_3=2\)，即 \(\displaystyle a_1\left(1+q+q^2\right)=2\)，即 \(\displaystyle a_1\left(1-q^3\right)=2\left(1-q\right)\)，整理得 \(\displaystyle a_1=2\left(q-1\right)\)。
考虑数列 \(\displaystyle a_n=a_1q^{n-1}=2q^n-2q^{n-1}\)，\(\displaystyle n=1,2,\cdots,9\)。则
\(\displaystyle a_1+a_2+\cdots+a_6=2\left(q^6-1\right)=6=2^2+2\)，
\(\displaystyle a_1+a_2+\cdots+a_9=2\left(q^9-1\right)=14\ne3^2+3\)。矛盾。
（2）如果 \(\displaystyle m=2\)，则 \(\displaystyle q=\sqrt[3]{\frac32}\)，
所以 \(\displaystyle a_4+a_5+a_6=4\)，即 \(\displaystyle a_4\left(1+q+q^2\right)=4\)，即 \(\displaystyle a_4\left(1-q^3\right)=4\left(1-q\right)\)，整理得 \(\displaystyle a_4=8\left(q-1\right)\)。
考虑数列 \(\displaystyle a_n=a_4q^{n-4}=8\left(q^{n-3}-q^{n-4}\right)\)，\(\displaystyle n=4,5,6,7,8,9\)。
则 \(\displaystyle a_4+a_5+a_6=8\left(q^3-1\right)=4\)，
\(\displaystyle a_7+a_8+a_9=8\left(q^6-q^3\right)=8\left(\frac94-\frac32\right)=6\)。
接下来随意构造即可，取 \(\displaystyle a_3=8\left(1-q^{-1}\right)\)，\(\displaystyle a_2=8\left(q^{-1}-q^{-2}\right)\)，\(\displaystyle a_1\) 使得 \(\displaystyle a_1+a_2+a_3=2\)；
取 \(\displaystyle a_{10}=8\left(q^7-q^6\right)\)，\(\displaystyle a_{11},a_{12}\) 使得 \(\displaystyle a_{10}+a_{11}+a_{12}=8\) 即可。
综上，\(\displaystyle q\) 的最大值是 \(\displaystyle \sqrt[3]{\frac32}\)。
注：条件ii可以加强为有连续的 \(\displaystyle 10\) 项成等比数列。
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