4.3 等比数列及其前n项和
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知识剖析
等比数列的定义
如果一个数列从第二项起,每一项与它的前一项的比等于同一个常数,那么这个数列叫做等比数列,这个常数叫做等比数列的公比,记为\(q\).
代数形式:\(\dfrac{a_{n}}{a_{n-1}}=q\)(\(q\)是常数,\(n≥2\)) 或\(\dfrac{a_{n+1}}{a_{n}}=q\)(\(q\)是常数,\(n∈ N^*\))
\({\color{Red}{Eg}}\) \(\dfrac{a_{n}}{a_{n-1}}=2(n \geq 2)\)\(\Rightarrow\left\{a_{n}\right\}\)是公比为\(2\)的等比数列;
\(\dfrac{a_{n+1}}{a_{n}}=-3\)\(\Rightarrow\left\{a_{n}\right\}\)是公比为\(-3\)的等比数列;
\(\dfrac{a_{n+1}}{a_{n}}=4 n\)\(\Rightarrow\left\{a_{n}\right\}\)不是等比数列;
\({\color{Red}{解释}}\)
所谓常数就是与\(n\)无关;等比数列中\(a_n≠0\),\(q≠0\);
偶数项的正负、奇数项的正负相同.
\({\color{Red}{Eg}}\) 若\(-1\),\(b_1\),\(b_2\),\(b_3\),\(-4\)成等比数列,则\(b_2=\)\(\underline{\quad \quad}\).
解:\(b_2^2=-1×(-4)=4⇒b_2=±2\),
而\(b_2=-1\cdot q^2<0\),故\(b_2=-2\).
\({\color{Red}{(b_2与-1,-4均是奇数项,符号相同)}}\)
等比中项
若\(a ,b ,c\)成等比数列,则\(b\)称\(a\)与\(c\)的等差中项,则\(b^2=ac\);
证明一个数列是等比数列的方法
① 定义法:\(\dfrac{a_{n}}{a_{n-1}}=q\)(\(q\)是常数,\(n≥2\)),则\(\{a_n\}\)是等比数列;
② 中项法:\(a_{n+1}^{2}=a_{n} a_{n+2}\)\((a_n≠ 0 ,n∈ N^*)\),则\(\{a_n\}\)是等比数列;
③ 通项公式法:若数列的通项公式是形如\(a_n=k\cdot q^n\)(\(k ,q\)是不为\(0\)常数),则数列\(\{a_n\}\)是等比数列;
④ 前\(n\)项和法:若数列的前\(n\)项和是形如\(S_n=k\cdot q^n-k\)(\(k ,q\)是常数且\(k≠0\),\(q≠0\),\(1\)),则数列\(\{a_n\}\)是等比数列.
通项公式
等比数列\(\{a_n\}\)的首项为\(a_1\),公比为\(q\),则\(a_n=a_1 q^{n-1}\)(由定义与累乘法可得)
前\(n\)项和
等比数列\(\{a_n\}\)的首项为\(a_1\),公比为\(q\),则其前\(n\)项和为
(由错位相减法可证)
\({\color{Red}{注}}\)使用时注意公比是否等于\(1\),若不确定,使用时需要分类讨论.
基本性质
(其中\(m ,n ,p ,t∈N^*\))
设\(\{a_n\}\)是首项为\(a_1\), 公比为\(q\)的等比数列,那么
\((1)\)若\(m+n=p+t\), 则\(a_m a_n=a_p a_t\);
\((2)\)\(a_n=a_m q^{n-m}\);
\((3)\)\(q^{n-m}=\dfrac{a_{n}}{a_{m}}\);
\((4)\)数列\(\left\{\lambda a_{n}\right\}\)(\(λ\)是不为零的常数)仍是公比为\(q\)的等比数列;若数列\(\{b_n\}\)是公比为\(t\)的等比数列,则数列\(\left\{a_{n} b_{n}\right\}\)是公比为\(q\cdot t\)的等比数列;
\((5)\)下标成等差数列且公差为\(m\)的项\(a_k\),\(a_{k+m}\),\(a_{k+2 m}\),…(\(k ,m∈N^*\))组成公比为\(q^m\)的等比数列;
\((6)\)若\(q≠-1\),则\(S_n\),\(S_{2n}-S_n\),\(S_{3n}-S_{2n}\),…成等比数列;
(\(∵q=-1\),\(n\)是偶数时,\(S_n=0\))
经典例题
【题型一】等比数列的判断与证明
【典题1】【多选题】已知数列\(\{a_n\}\)是等比数列,那么下列数列一定是等比数列的是( )
A.\(\left\{\dfrac{1}{a_{n}}\right\}\) \(\qquad \qquad \qquad \qquad\)B.\(\left\{\log _{2} a_{n}\right\}\)\(\qquad \qquad \qquad \qquad\)C.\(\left\{a_{n} a_{n+1}\right\}\) \(\qquad \qquad \qquad \qquad\)D.\(\left\{a_{n}+a_{n+1}+a_{n+2}\right\}\)
【解析】由题意,可设等比数列\(\{a_n\}\)的公比为\(q(q≠0)\),则\(a_n=a_1 q^{n-1}\).
对于\(A\):\(\dfrac{1}{a_{n}}=\dfrac{1}{a_{1} q^{n-1}}=\dfrac{1}{a_{1}}\left(\dfrac{1}{q}\right)^{n-1}\).
(等比数列通项公式形如指数型\(a_n=A\cdot B^n\))
\(∴\)数列\(\left\{\dfrac{1}{a_{n}}\right\}\)是一个以\(\dfrac{1}{a_{1}}\)为首项,\(\dfrac{1}{q}\)为公比的等比数列;
对于\(B\):若\(a_n>0\),则\(\log _{2} a_{n}=\log _{2}\left(a_{1} q^{n-1}\right)\)\(=\log _{2} a_{1}+(n-1) \log _{2} q\)
\(∴\)数列\(\left\{\log _{2} a_{n}\right\}\)是一个以\(\log _{2} a_{1}\)为首项,\(\log _{2} q\)为公差的等差数列;
对于\(C\):\(\because \dfrac{a_{n+1} \cdot a_{n+2}}{a_{n} \cdot a_{n+1}}=\dfrac{a_{n+2}}{a_{n}}=\dfrac{a_{1} \cdot q^{n+1}}{a_{1} \cdot q^{n-1}}=q^{2}\)
\(∴\)数列\(\left\{a_{n}+a_{n+1}+a_{n+2}\right\}\)是一个以\(q^2\)为公比的等比数列
对于\(D\):\(\because \dfrac{a_{n+1}+a_{n+2}+a_{n+3}}{a_{n}+a_{n+1}+a_{n+2}}\)\(=\dfrac{q\left(a_{n}+a_{n+1}+a_{n+2}\right)}{a_{n}+a_{n+1}+a_{n+2}}=q\)
\(∴\)数列\(\left\{a_{n}+a_{n+1}+a_{n+2}\right\}\)是一个以\(q\)为公比的等比数列.
故选:\(ACD\).
【点拨】
① 判定等比数列常用定义法:\(\dfrac{a_{n+1}}{a_{n}}=q\)(\(q\)是常数,\(n∈ N^*\))\(⇒\{a_n\}\)是等比数列;
② 等比数列通项公式形如指数型\(a_n=k\cdot q^n\),在选择填空题运用.
【典题2】已知数列\(\{a_n\}\)的前\(n\)项和为\(S_n\),且满足\(a_{n+1}=S_{n}+n+1\)\(\left(n \in N^{*}\right)\),\(a_1=1\).
求证\(\left\{a_{n}+1\right\}\)为等比数列,并求\(a_n\).
【解析】证明:\(\because a_{n+1}=S_{n}+n+1\left(n \in N^{*}\right)\),
\({\color{Red}{(遇到a_n与S_n的等式可想到a_{n}=\left\{\begin{array}{cc}
S_{1}, & n=1 \\
S_{n}-S_{n-1}, & n \geq 2
\end{array}\right. )}}\)
当\(n≥2\)时,\(a_{n}=S_{n-1}+n\)
两式相减得\(a_{n+1}-a_{n}=S_{n}+n+1-\left(S_{n-1}+n\right)=a_{n}+1\)
\(\therefore a_{n+1}+1=2 a_{n}+2=2\)\(\left(a_{n}+1\right)(n \geq 2)\),
\({\color{Red}{(不要漏了大前提:n≥2,要证明\{a_{n}+1\}为等比数列,还要判断a_{n+1}+1=2(a_{n}+1)当n=1时也成立)}}\)
而对于\(a_{n+1}=S_{n}+n+1\left(n \in N^{*}\right)\)中
当\(n=1\)时有\(a_2=S_1+2=a_1+2=3\),
则\(a_2=3\),\(a_1=1\)满足\(a_{n+1}+1=2\left(a_{n}+1\right)\),
\(\therefore a_{n+1}+1=2\left(a_{n}+1\right)\)\(\left(n \in N^{*}\right)\),
\(\therefore\left\{a_{n}+1\right\}\)为等比数列,首项为\(a_1+1=2\),公比为\(2\)的等比数列,
\(\therefore a_{n}+1=2 \cdot 2^{n-1}=2^{n}\),
\(∴a_n=2^n-1\).
【点拨】数列问题中,特别要注意\(n\)的取值范围,比如\(n≥2\)、\(n∈N^*\),要确定好.
【典题3】设数列\(\{a_n\}\)的首项\(a_1\)为常数,且\(a_{n}=3^{n-1}-2 a_{n-1}\)\((n≥2)\).
(1) 判断数列\(\left\{a_{n}-\dfrac{3^{n}}{5}\right\}\)是否为等比数列,请说明理由;
(2)\(S_n\)是数列\(\{a_n\}\)的前\(n\)项的和,若\(\left\{S_{n}\right\}\)是递增数列,求\(a_1\)的取值范围.
【解析】(1)当\(n≥2\)时,\(a_{n}-\dfrac{3^{n}}{5}=3^{n-1}-2 a_{n-1}-\dfrac{3^{n}}{5}\)\(=2 \cdot \dfrac{3^{n-1}}{5}-2 a_{n-1}=-2\left(a_{n-1}-\dfrac{3^{n-1}}{5}\right)\)
\({\color{Red}{(定义法证明等比数列,要注意首项a_{1}-\dfrac{3}{5}是否等于0)}}\)
① 当\(a_{1}-\dfrac{3}{5} \neq 0\),即\(a_{1} \neq \dfrac{3}{5}\)时,\(\dfrac{a_{n}-\dfrac{3^{n}}{5}}{a_{n-1}-\dfrac{3^{n-1}}{5}}=-2\),
\(\therefore a_{1} \neq \dfrac{3}{5}\)时,\(\left\{a_{n}-\dfrac{3^{n}}{5}\right\}\)为等比数列,公比为\(-2\).
② 当\(a_{1}=\dfrac{3}{5}\),即\(a_{1}=\dfrac{3}{5}\)时,\(a_{n}-\dfrac{3^{n}}{5}=0\),数列\(\left\{a_{n}-\dfrac{3^{n}}{5}\right\}\)不是等比数列.
(2) ① 当\(a_{1}=\dfrac{3}{5}\)时,\(a_{n}=\dfrac{1}{5} \cdot 3^{n}\),为单调递增数列,满足条件.
② 当\(a_{1} \neq \dfrac{3}{5}\)时,由(1)可得:\(a_{n}-\dfrac{3^{n}}{5}=\left(a_{1}-\dfrac{3}{5}\right)(-2)^{n-1}\),
若\(\left\{S_{n}\right\}\)是递增数列,
则\(S_{n}-S_{n-1}>0(n \geq 2)\),即\(a_n>0(n≥2)\),
\(\therefore a_{n}=\left(a_{1}-\dfrac{3}{5}\right)(-2)^{n-1}+\dfrac{3^{n}}{5}>0\),
\({\color{Red}{(问题变成恒成立问题,可想到分离参数法,遇到(-2)^{n-1}想到分n奇偶数讨论)}}\)
当\(n\)为偶数,则\(-2^{n-1}\left(a_{1}-\dfrac{3}{5}\right)+\dfrac{3^{n}}{5}>0 \Rightarrow a_{1}<\dfrac{2}{5} \cdot\left(\dfrac{3}{2}\right)^{n}+\dfrac{3}{5}\),
\(\therefore a_{1}<\dfrac{2}{5} \cdot\left(\dfrac{3}{2}\right)^{2}+\dfrac{3}{5}=\dfrac{3}{2}\),
\({\color{Red}{(f(n)=\dfrac{2}{5} \cdot\left(\dfrac{3}{2}\right)^{n}+\dfrac{3}{5}是增数列,f_{\min }=f(2))}}\)
当\(n\)为奇数,则\(2^{n-1}\left(a_{1}-\dfrac{3}{5}\right)+\dfrac{3^{n}}{5}>0\)\(\Rightarrow a_{1}>-\dfrac{2}{5} \cdot\left(\dfrac{3}{2}\right)^{n}+\dfrac{3}{5}\),
故\(\text { 籹 } a_{1}>-\dfrac{2}{5} \cdot\left(\dfrac{3}{2}\right)^{3}+\dfrac{3}{5}=-\dfrac{3}{4}\),
\({\color{Red}{(f(n)=-\dfrac{2}{5} \cdot\left(\dfrac{3}{2}\right)^{n}+\dfrac{3}{5}是减数列,f_{\max }=f(3))}}\)
\(\therefore-\dfrac{3}{4}<a_{1}<\dfrac{3}{2}\).且\(a_{1} \neq \dfrac{3}{5}\).
综上可得:\(-\dfrac{3}{4}<a_{1}<\dfrac{3}{2}\).
【点拨】若\(a_{n+1}=2 a_{n}\)\(\left(n \in N^{*}\right)\),不能得到\(\{a_n\}\)是等比数列,一定要强调\(a_1≠0\)才行.
巩固练习
1(★)根据下列通项能判断数列为等比数列的是( )
A.\(a_n=n\) \(\qquad \qquad \qquad \qquad\) B.\(a_{n}=\sqrt{n}\) \(\qquad \qquad \qquad \qquad\)C.\(a_{n}=2^{-n}\) \(\qquad \qquad \qquad \qquad\)D.\(a_{n}=\log _{2} n\)
2(★★)已知数列\(\{a_n\}\)是等比数列,则下列数列中:①\(\left\{a_{n}^{3}\right\}\);②\(\left\{2^{a_{n}}\right\}\);③\(\left\{\dfrac{1}{2 a_{n}}\right\}\),等比数列的个数是( )
A.\(0\)个 \(\qquad \qquad \qquad \qquad\)B.\(1\)个 \(\qquad \qquad \qquad \qquad\)C.\(2\)个 \(\qquad \qquad \qquad \qquad\)D.\(3\)个
3(★★)在数列\(\{a_n\}\)中,\(a_1=2\),\(a_{n+1}=4 a_{n}-3 n+1\),\(n∈N^*\),证明:数列\(\left\{a_{n}-n\right\}\)是等比数列.
4(★★★)设\(S_n\)为数列\(\{a_n\}\)的前\(n\)项和,已知\(a_3=7\),\(a_{n}=2 a_{n-1}+a_{2}-2\)\((n \geq 2)\).
(1)证明:\(\left\{a_{n}+1\right\}\)为等比数列;
(2)求\(\{a_n\}\)的通项公式,并判断\(n\),\(a_n\),\(S_n\)是否成等差数列?
答案
1.\(C\)
2.\(C\)
3. 提示:定义法证明
4.\((1)\)提示:定义法证明
\((2) a_n=2^n-1\),\(n\),\(a_n\),\(S_n\)成等差数列
【题型二】等比数列的基本运算
【典题1】若\(S_n\)是等比数列\(\{a_n\}\)的前n项和,\(S_3\),\(S_9\),\(S_6\)成等差数列,且\(a_7=2\),则\(a_4+a_{10}=\) \(\underline{\quad \quad}\).
【解析】由题意可得:等比数列\(\{a_n\}\)的公比\(q≠1\),
\({\color{Red}{(利用等比数列的前n项和公式S_{n}= \begin{cases}n a_{1} & (q=1) \\ \dfrac{a_{1}\left(1-q^{n}\right)}{1-q} & (q \neq 1)\end{cases},特别要注意公比q是否等于1)}}\)
\(∵S_3\),\(S_9\),\(S_6\)成等差数列,\(∴2S_9=S_6+S_3,\)
\(\therefore 2 \times \dfrac{a_{1}\left(q^{9}-1\right)}{q-1}=\dfrac{a_{1}\left(q^{6}-1\right)}{q-1}+\dfrac{a_{1}\left(q^{3}-1\right)}{q-1}\)\(\Rightarrow 2 q^{6}-q^{3}-1=0 \Rightarrow\left(2 q^{3}+1\right)\left(q^{3}-1\right)=0\),
\(\therefore q^{3}=-\dfrac{1}{2}\)
\(\because a_{7}=2 \quad \therefore a_{1} q^{6}=2 \Rightarrow a_{1}=8\),
则\(a_4+a_{10}=a_1 q^3+a_1 q^9= -4-1=-5\).
【点拨】
① 与等差数列差不多,首项\(a_1\)和公比\(q\)是等比数列的基本量,通项公式\(a_{n}=a_{1} q^{n-1}\)和前\(n\)项和公式\(S_{n}= \begin{cases}n a_{1} & (q=1) \\ \dfrac{a_{1}\left(1-q^{n}\right)}{1-q} & (q \neq 1)\end{cases}\)均与基本量有关;
②等比数列中\(a_1\),\(q\),\(a_n\),\(S_n\),\(n\)五个量,一般可以“知二求三”,通过条件得到\(a_1\)与\(q\)的方程(组)是关键.
【典题2】【多选题】正项等比数列\(\{a_n\}\)的前\(n\)项和是\(S_n\),已知\(S_3=a_2+10a_1\),\(a_4=3\).下列说法正确的是( )
A.\(a_1=9\)
B.\(\{a_n\}\)是递增数列
C.\(\left\{S_{n}+\dfrac{1}{18}\right\}\)为等比数列
D.\(\left\{\log _{3} a_{n}\right\}\)是等比数列
【解析】\(∵S_3=a_2+10a_1\),\(a_4=3\).
设首项为\(a_1\),公比为\(q\),
则\(\left\{\begin{array}{c}
a_{1}+a_{2}+a_{3}=a_{2}+10 a_{1} \\
a_{1} q^{3}=3
\end{array}\right.\),
解得\(\left\{\begin{array}{l}
a_{1}=\dfrac{1}{9} \\
q=3
\end{array}\right.\),
\({\color{Red}{(这里S_3=a_1+a_2+a_3,没用等比数列前n项和公式,运算简单些)}}\)
所以\(a_{n}=\dfrac{1}{9} \cdot 3^{n-1}=3^{n-3}\),故\(A\)错误,\(B\)正确;
则\(S_{n}=\dfrac{\dfrac{1}{9}\left(3^{n}-1\right)}{3-1}=\dfrac{1}{18} \cdot 3^{n}-\dfrac{1}{18}\),
由于\(S_n\)的关系式符合\(kq^n-k\)的形式,故\(C\)正确.
由于\(a_{n}=3^{n-3}\),
所以\(\log _{3} a_{n}=\log _{3} 3^{n-3}=n-3\),
所以该数列为等差数列,故\(D\)错误.
故选:\(BC\).
【点拨】
① 在等比数列中遇到\(S_n\)若其下标较小,用\(S_n=a_1+a_2+⋯+a_n\)会比等比数列的前\(n\)项和公式更好些;
② 本题还是把已知条件向基本量“靠拢”.
【典题3】数列\(\{a_n\}\)、\(\{b_n\}\)均为等比数列,其前\(n\)项和分别为\(S_n\),\(T_n\),若对任意的\(n∈N^*\),都有\(\dfrac{S_{n}}{T_{n}}=\dfrac{3^{n}+1}{4}\),则\(\dfrac{a_{4}}{b_{4}}=\) \(\underline{\quad \quad}\).
【解析】当\(n=1\)时,\(\dfrac{S_{1}}{T_{1}}=\dfrac{3+1}{4}=1\),
即\(a_1=b_1\)
设\(\{a_n\}\)、\(\{b_n\}\)的公比分别为\(q\),\(p\),
则\(\dfrac{S_{2}}{T_{2}}=\dfrac{a_{1}+a_{1} q}{b_{1}+b_{1} p}=\dfrac{1+q}{1+p}=\dfrac{9+1}{4}=\dfrac{5}{2}\),
即\(2(1+q)=5(1+p)\),即\(q=\dfrac{3}{2}+\dfrac{5 p}{2}\),
当\(n=3\)时,\(\dfrac{S_{3}}{T_{3}}=\dfrac{a_{1}+a_{1} q+a_{1} q^{2}}{b_{1}+b_{1} p+b_{1} p^{2}}=\dfrac{1+q+q^{2}}{1+p+p^{2}}=\dfrac{27+1}{4}=7\)
即\(1+q+q^2=7(1+p+p^2)\),
将\(q=\dfrac{3}{2}+\dfrac{5 p}{2}\)代入得\(1+\dfrac{3}{2}+\dfrac{5 p}{2}+\left(\dfrac{3}{2}+\dfrac{5 p}{2}\right)^{2}=7\left(1+p+p^{2}\right)\),
整理得\(p^2-4p+3=0\),得\(p=1\)或\(3\),
当\(p=1\)时,\(q=\dfrac{3}{2}+\dfrac{5 p}{2}=4\),
此时\(\dfrac{S_{n}}{T_{n}}=\dfrac{\dfrac{a_{1}\left(1-4^{n}\right)}{1-4}}{n b_{1}}=\dfrac{4^{n}-1}{3 n} \neq \dfrac{3^{n}+1}{4}\)
\(∴p=1\)不成立,
当\(p=3\)时,\(q=9\),此时\(\dfrac{a_{4}}{b_{4}}=\dfrac{a_{1} q^{3}}{b_{1} p^{3}}=\dfrac{9^{3}}{3^{3}}=27\)
综上\(\dfrac{a_{4}}{b_{4}}=27\).
【点拨】利用方程思想向基本量靠拢进行求解,计算量略大.
巩固练习
1(★★)已知数列\(\{a_n\}\)中,\(a_1=7\),\(a_3=1\),若\(\left\{\dfrac{1}{a_{n}+1}\right\}\)是等比数列,则\(a_{11}\)等于\(\underline{\quad \quad}\).
2(★★)已知数列\(\{a_n\}\)是各项均为正数的等比数列,其前\(n\)项和为\(S_n\),\(a_1+a_2=2\),\(a_5+a_6=8\),则\(S_{10}=\) \(\underline{\quad \quad}\).
3(★★)已知等比数列\(\{a_n\}\)的前\(n\)项和是\(S_n\),若\(a_1+2a_2=0\),\(S_{3}=\dfrac{3}{4}\),且\(a≤S_n≤a+2\),则实数\(a\)的取值范围是\(\underline{\quad \quad}\).
4(★★)若等比数列\(\{a_n\}\)的前\(n\)项和是\(S_n\),且\(S_2=3\),\(S_6=63\),则\(S_5=\) \(\underline{\quad \quad}\).
5(★★)设\(\{a_n\}\)是各项均为正数的等比数列,\(S_n\)为其前\(n\)项和.已知\(a_1 a_3=16\),\(S_3=14\),若存在\(n_0\)使得\(a_1\),\(a_2\),⋯ ,\(a_{n_0}\)的乘积最大,则\(n_0\)的一个可能值是( )
A.\(4\) \(\qquad \qquad \qquad \qquad\)B.\(5\) \(\qquad \qquad \qquad \qquad\)C.\(6\) \(\qquad \qquad \qquad \qquad\)D.\(7\)
6(★★★)【多选题】在公比\(q\)为整数的等比数列\(\{a_n\}\)中,\(S_n\)是数列\(\{a_n\}\)的前\(n\)项和,若\(a_1 a_4=32\),\(a_2+a_3=12\),则下列说法正确的是( )
A.\(q=2\) \(\qquad \qquad \qquad \qquad \qquad \qquad\) B.数列\(\left\{S_{n}+2\right\}\)是等比数列
C.\(S_8=510\) \(\qquad \qquad \qquad\qquad \qquad \qquad\) D.数列\(\left\{\lg a_{n}\right\}\)是公差为\(2\)的等差数列
7(★★★)【多选题】已知等比数列\(\{a_n\}\)公比为\(q\),前\(n\)项和是\(S_n\),且满足\(a_6=8a_3\),则下列说法正确的是( )
A.\(q=2\) \(\qquad \qquad \qquad \qquad\)B.\(\dfrac{S_{6}}{S_{3}}=9\)\(\qquad \qquad \qquad \qquad\) C.\(S_3\),\(S_6\),\(S_9\)成等比数列 \(\qquad \qquad \qquad \qquad\)D.\(S_n=2a_n+a_1\)
参考答案
1.\(-\dfrac{127}{128}\)
2.\(62\)
3.\(\left[-1, \dfrac{1}{2}\right]\)
4.\(-33\)或\(31\)
5.\(A\)
6.\(ABC\)
7.\(AB\)
【题型三】等比数列的基本性质及运用
【典题1】已知等比数列\(\{a_n\}\)的公比大于\(1\),\(a_3 a_7=72\),\(a_2+a_8=27\),则\(a_{12}=\) \(\underline{\quad \quad}\).
【解析】在公比大于\(1\)的等比数列\(\{a_n\}\)中,
\(∵a_3 a_7=72⇒a_2 a_8=27\),\(a_2+a_8=27\)
则\(a_2\),\(a_8\)是\(x^2-27x+72=0\)的两根,
\({\color{Red}{(利用韦达定理,求解更快捷)}}\)
可解得\(\left\{\begin{array}{c}
a_{2}=3 \\
a_{8}=24
\end{array}\right.\)或\(\left\{\begin{array}{c}
a_{8}=3 \\
a_{2}=24
\end{array}\right.\),
由于公比大于\(1\),则\(a_2=3\),\(a_8=24\),
则有\(q^{6}=\dfrac{a_{8}}{a_{2}}=8\),则\(q^2=2\),
\({\color{Red}{(q^{n-m}=\dfrac{a_{n}}{a_{m}},知道两项便可直接求出公比q)}}\)
\(a_{12}=a_2 q^{10}=3×2^5=96\).
\({\color{Red}{(a_{n}=a_{m} q^{n-m},求任意一项a_k不一定要知道a_1)}}\)
【点拨】本题若使用方程思想求基本量的方法,计算量就较大;注意项数的下标之间数值的关系,利用等比数列的相关性质求解快捷.
【典题2】【多选题】已知等比数列\(\{a_n\}\)的各项均为正数,公比为\(q\),且\(a_1>1\),\(a_6+a_7>a_5 a_8+1>2\),记\(\{a_n\}\)的前\(n\)项积为\(T_n\),则下列选项中正确的选项是 ( )
A.\(0<q<1\) \(\qquad \qquad \qquad \qquad\)B.\(a_6>1\) \(\qquad \qquad \qquad \qquad\)C.\(T_{12}>1\) \(\qquad \qquad \qquad \qquad\) D.\(T_{13}>1\)
【解析】\(∵a_6+a_7>a_5 a_8+1\)
\(∴a_6+a_7>a_6 a_7+1\)\(⇒a_6 a_7-a_6-a_7+1<0\)
\(∴(a_6-1)(a_7-1)<0 (*)\),
\(∵a_1>1\),
若\(a_6<1⇒a_1 q^5<1⇒0<q<1\),
则一定有\(a_7=a_6 q<1\),不符合 (*),
\({\color{Red}{(大胆假设小心验证)}}\)
则\(a_6>1 ,a_7<1\),
\(∴0<q<1\).
\(∵a_6 a_7+1>2\),
\(∴a_6 a_7>1\),
\(∴T_{12}=a_1 a_2 a_3…a_{12}=(a_6 a_7 )^6>1\),\(T_{13}=a_{1} a_{2} a_{3} \ldots a_{12} a_{13}=a_{7}^{13}<1\),
故选:\(ABC\).
【点拨】注意到题中\(a_6\)、\(a_7\)、\(a_5\)、\(a_8\)下标存在\(6+7=5+8\),而\(T_{12} 、 T_{13}\)与\(a_6\)、\(a_7\)有关,故利用等比数列的性质:若\(m+n=p+t\), 则\(a_m a_n=a_p a_t\).
【典题3】已知正项等比数列\(\{a_n\}\)的前\(n\)项和为\(S_n\)且\(S_8-2S_4=6\),则\(a_9+a_{10}+a_{11}+a_{12}\)的最小值为\(\underline{\quad \quad}\).
【解析】由于\(a_n>0\),则公比\(q\)不可能等于\(-1\),
由等比数列的性质可得:\(S_4\),\(S_8-S_4\),\(S_{12}-S_8\)成等比数列,
则\(S_{4}\left(S_{12}-S_{8}\right)=\left(S_{8}-S_{4}\right)^{2}\)\(\Rightarrow S_{12}-S_{8}=\dfrac{\left(S_{8}-S_{4}\right)^{2}}{S_{4}}\)\(=\dfrac{\left(S_{4}+6\right)^{2}}{S_{4}}=S_{4}+\dfrac{36}{S_{4}}+12\),
\(\therefore a_{9}+a_{10}+a_{11}+a_{12}\)\(=S_{12}-S_{8}=S_{4}+\dfrac{36}{S_{4}}+12 \geq 24\),
当且仅当\(S_4=6\)时等号成立.
则\(a_9+a_{10}+a_{11}+a_{12}\)的最小值为\(24\).
【点拨】若\(q≠-1\),则\(S_{n}\),\(S_{2 n}-S_{n}\),\(S_{3 n}-S_{2 n}\),…成等比数列.
巩固练习
1(★★)若等比数列\(\{a_n\}\)的各项均为正数,且\(a_1 a_{10}=9\),则\(\log _{9} a_{1}+\log _{9} a_{2}+\cdots+\log _{9} a_{10}=\) \(\underline{\quad \quad}\).
2(★★)正项等比数列\(\{a_n\}\)满足\(a_2^2+2a_3a_7\)\(+a_6 a_{10}=16\),则\(a_2+a_8=\) \(\underline{\quad \quad}\).
3(★★)在等比数列\(\{a_n\}\)中,若\(a_{2} a_{5}=-\dfrac{3}{4}\),\(a_{2}+a_{3}+a_{4}+a_{5}=\dfrac{9}{4}\),则\(\dfrac{1}{a_{2}}+\dfrac{1}{a_{3}}+\dfrac{1}{a_{4}}+\dfrac{1}{a_{5}}=\) \(\underline{\quad \quad}\).
4(★★)已知等比数列\(\{a_n\}\)中,\(a_2 a_3 a_4=1\),\(a_6 a_7 a_8=64\),则\(a_4 a_5 a_6=\) \(\underline{\quad \quad}\).
5(★★)等比数列\(\{a_n\}\)中,\(a_1+a_2+a_3=3\),\(a_4+a_5+a_6=6\),则\(\{a_n\}\)的前\(12\)项和为\(\underline{\quad \quad}\).
6(★★)若等比数列\(\{a_n\}\)的前\(n\)项和为\(S_n\),且\(\dfrac{S_{6}}{S_{3}}=6\),则\(\dfrac{S_{9}}{S_{6}}=\) \(\underline{\quad \quad}\).
7(★★)已知等比数列\(\{a_n\}\)的前\(n\)项和为\(S_n\),若\(a_1+a_3=5\),\(S_4=20\),则\(\dfrac{S_{8}-2 S_{4}}{S_{6}-S_{4}-S_{2}}=\) \(\underline{\quad \quad}\).
8(★★)已知正项等比数列\(\{a_n\}\)满足\(a_2 a_8=16a_5\),\(a_3+a_5=20\),若存在两项\(a_m\),\(a_n\)使得\(\sqrt{a_{m} a_{n}}=32\),则\(\dfrac{1}{m}+\dfrac{4}{n}\)的最小值为\(\underline{\quad \quad}\).
9(★★)【多选题】设等比数列\(\{a_n\}\)的公比为\(q\),其前\(n\)项和是\(S_n\),前\(n\)项积为\(T_n\),并且满足条件\(a_1>1\),\(a_7 a_8>1\),\(\dfrac{a_{7}-1}{a_{8}-1}<0\),则下列结论正确的是( )
A.\(0<q<1\) \(\qquad \qquad \qquad\)B.\(a_7 a_9>1\) \(\qquad \qquad \qquad\)C.\(S_n\)的最大值为\(S_9\) \(\qquad \qquad \qquad\)D.\(T_n\)的最大值为\(T_7\)
10(★★)【多选题】等比数列\(\{a_n\}\)的公比为\(q\),且满足\(a_1>1\),\(a_{1010}a_{1011}>1\),\(\left(a_{1010}-1\right)\left(a_{1011}-1\right)<0\),记\(T_n=a_1 a_2 a_3…a_n\),则下列结论正确的是( )
A.\(0<q<1\) \(\qquad\) B.\(a_{1010} a_{1012}-1>0\) \(\qquad\)C.\(T_{n} \leq T_{1011}\) \(\qquad\) D.使\(T_n<1\)成立的最小自然数\(n\)等于\(2021\)
参考答案
1.\(5\)
2.\(4\)
3.\(-3\)
4.\(8\)
5.\(45\)
6.\(\dfrac{31}{6}\)
7.\(10\)
8.\(\dfrac{3}{4}\)
9.\(AD\)
10.\(AD\)