4.3.1 等比数列的概念1(概念、通项公式)
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选择性第二册同步巩固,难度2颗星!
基础知识
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,\(n∈ N^* )\)
解释
(1) 公比是每一项与它的前一项的比,常数指的是与\(n\)无关;
(2) 等比数列中\(a_n≠0\) ,\(q≠0\)(否则数列会出现\(0\),不可能符合等边数列定义);
(3) \(\dfrac{a_n}{a_{n-1}}=2(n \geq 2) \Rightarrow\left\{a_n\right\}\)是公比为\(2\)的等比数列;
\(\qquad\)\(\dfrac{a_{n+1}}{a_n}=-3 \Rightarrow\left\{a_n\right\}\)是公比为\(-3\)的等比数列;
\(\qquad\)\(\dfrac{a_{n+1}}{a_n}=4 n \Rightarrow\left\{a_n\right\}\)不是等比数列.
【例】 以下数列是等比数列的\(\underline{\quad \quad}\).
(1)\(2,4,8,16\);\(\qquad \qquad\) (2)\(a,a,a,a\);\(\qquad \qquad\) (3)数列\(\{a_n \}\)满足 \(\dfrac{a_n}{a_{n-1}}=-2(n \geq 2)\).
答案 (1)(3).当\(a=0\)时,(2)不是等比数列
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 若\(a\),\(b\) ,\(c\)成等比数列,由等比数列的定义可得\(\dfrac{b}{a}=\dfrac{c}{b}\),即\(b^2=ac\).
【例】 若\(2x\)是\(1\)与\(x+3\)的等比中项,则\(x=\)\(\underline{\quad \quad}\) .
解 依题意得\(4x^2=1\cdot (x+3)\),解得\(x=1\)或 \(-\dfrac{3}{4}\).
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)
解释
(1) 证明 由等比数列的定义可得, \(\dfrac{a_{n+1}}{a_n}=q\),
所以\(\dfrac{a_2}{a_1}=q\), \(\dfrac{a_3}{a_2}=q\), \(\dfrac{a_4}{a_3}=q\),\(…\)\(\dfrac{a_n}{a_{n-1}}=q(n \geq 2)\),
把以上\(n-1\)个等式累乘可得\(\dfrac{a_n}{a_1}=q^{n-1}(n \geq 2)\),即\(a_n=a_1 q^{n-1} (n≥2)\),
当\(n=1\)时,\(a_1=a_1 q^0=a_1 q^{1-1}\),即当\(n=1\)时上式也成立,
故\(a_n=a_1 q^{n-1} (n∈N^*)\).
以上的方法称之为累乘法 .
(2) 通项公式\(a_n=a_1 q^{n-1}\)告诉你:已知等比数列的首项\(a_1\)与公比\(q\)可求得任何一项
(3)等比数列的通项公式可整理为\(a_n=a_1 q^{n-1}=\dfrac{a_1}{q} \cdot q^n\),当\(q>0\),且\(q≠1\)时,可以看成\(n\)的指数函数型函数.
比如等比数列\(\{2^n\}\)的各点都在指数函数\(y=2^x\)上.
(4)偶数项的正负、奇数项的正负相同(\(\dfrac{a_{2 n}}{a_{2(n-1)}}= q^2>0\),故\(a_{2n}\),\(a_{2(n-1)}\)同号,即偶数项的正负相同;奇数项同理).
【例1】 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\(b_2=-2\).
(\(b_2\)与\(-1\) ,\(-4\)均是奇数项,符号相同\()\)
【例2】 等比数列\(\{a_n \}\)中,\(a_1=2\),\(q=3\),则\(a_n\)等于\(\underline{\quad \quad}\) .
答案 \(2×3^{n-1}\)
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,\(n≥2)⇒\{a_n\}\)是等比数列;
② 中项法:\(a_{n+1}^2=a_n a_{n+2} (a_n≠ 0 ,n∈ N^*)⇒\{a_n\}\)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.
基本方法
【题型1】等比数列的判定和证明
【典题1】 (多选)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.\(\left\{\dfrac{1}{a_n}\right\}\) \(\qquad \qquad \qquad\) B.\(\{ \log_2a_n \}\) \(\qquad \qquad \qquad\) C.\(\{a_n\cdot a_{n+1}\}\) \(\qquad \qquad \qquad\) D.\(\{a_n+a_{n+1}+a_{n+2}\}\)
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\(A\): \(\dfrac{1}{a_n}=\dfrac{1}{a_1 q^{n-1}}=\dfrac{1}{a_1}\left(\dfrac{1}{q}\right)^{n-1}\).
\(\therefore\)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\(B\): \(\log _2 a_n=\log _2\left(a_1 q^{n-1}\right)=\log _2 a_1+(n-1) \log _2 q\).
\(\therefore\)数列\(\{ \log_2a_n \}\)是一个以\(\log_2a_1\)为首项,\(\log_2q\)为公差的等差数列;
对于\(C\): \(\because \dfrac{a_{n+1} \cdot a_{n+2}}{a_n \cdot a_{n+1}}=\dfrac{a_{n+2}}{a_n}=\dfrac{a_1 \cdot q^{n+1}}{a_1 \cdot q^{n-1}}=q^2\),
\(\therefore\)数列\(\{a_n a_{n+1}\}\)是一个以\(q^2\)为公比的等比数列;
对于\(D\): \(\because \dfrac{a_{n+1}+a_{n+2}+a_{n+3}}{a_n+a_{n+1}+a_{n+2}}=\dfrac{q\left(a_n+a_{n+1}+a_{n+2}\right)}{a_n+a_{n+1}+a_{n+2}}=q\),
\(\therefore\)数列\(\{a_n+a_{n+1}+a_{n+2}\}\)是一个以\(q\)为公比的等比数列.
故选:\(ACD\).
点拨 证明\(\{a_n \}\)是等比数列常见的方法是定义法: \(\dfrac{a_n}{a_{n-1}}=q\)\(( q\)是常数,\(n≥2)\);
选择题也可采取排除法,检验前三项是否成等比数列.
【典题2】已知数列\(\{a_n \}\)的前\(n\)项和为\(S_n\),且满足 \(S_n=\dfrac{3}{2} a_n+b\)\(\left(n \in N^*, b \in R, b \neq 0\right)\).
(1)求证:\(\{a_n \}\)是等比数列;\(\qquad \qquad\) (2)求证:\(\{a_n+1\}\)不是等比数列.
证明 (1)因为 \(S_n=\dfrac{3}{2} a_n+b\),所以当\(n≥2\)时 \(S_{n-1}=\dfrac{3}{2} a_{n-1}+b\),
两式相减得 \(S_n-S_{n-1}=\dfrac{3}{2} a_n+b-\dfrac{3}{2} a_{n-1}-b\),
\(\therefore a_n=\dfrac{3}{2} a_n-\dfrac{3}{2} a_{n-1}\),
\(\therefore a_n=3a_{n-1}\),
故\(\{a_n \}\)是公比为\(q=3\)的等比数列.
(2)假设:\(\{a_n+1\}\)是等比数列,则有:\((a_n+1)^2=(a_{n+1}+1)( a_{n-1}+1)\),
即:\(a_n^2+2a_n+1=a_{n+1} a_{n-1}+a_{n+1}+a_{n-1}+1\),
由(1)知\(\{a_n \}\)是等比数列,所以\(a_n^2=a_{n+1} a_{n-1}\),
于是\(2a_n=a_{n+1}+a_{n-1}\),即\(6a_n=a_{n-1}+9a_{n-1}\),解得\(a_{n-1}=0\),
这与\(\{a_n \}\)是等比数列相矛盾,
故假设错误,即:\(\{a_n+1\}\)不是等比数列.
【巩固练习】
1.根据下列通项能判断数列为等比数列的是( )
A.\(a_n=n\) \(\qquad \qquad \qquad\) B. \(a_n=\sqrt{n}\) \(\qquad \qquad \qquad\) C.\(a_n=2^{-n}\) \(\qquad \qquad \qquad\) D.\(a_n= \log_2n\)
2.已知数列\(\{a_n \}\)是等比数列,则下列数列中:①\(\{a_n^3\}\);②\(\{2^{a_n}\}\);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.已知数列\(\{a_n \}\)满足\(\lga_n=3n+5\),求证:\(\{a_n \}\)是等比数列.
4.已知数列\(\{a_n \}\)满足\(a_1=1\),\(2a_{n+1}=3a_n+1\).证明:\(\{a_n+1\}\)是等比数列.
参考答案
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答案 \(C\)
解析 在\(A\)中,\(a_n=n\)是等差数列,不是等比数列,故\(A\)错误;
在\(B\)中, \(a_n=\sqrt{n}\)既不是等差数列,又不是等比数列,故\(B\)错误;
在\(C\)中,\(a_n=2^{-n}\)是等比数列,故\(C\)正确;
在\(D\)中,\(a_n= \log_2n\)既不是等差数列,又不是等比数列,故\(D\)错误.
故选:\(C\). -
答案 \(C\)
解析 \(\because\)数列\(\{a_n \}\)是等比数列,设公比为\(q\),则\(q\)为常数.
则 \(a_n^3=a_1^3 q^{3 n-3}\),则 \(\dfrac{a_{n+1^3}}{a_n^3}=q^3\),为常数,故:①\(\{a_n^3\}\)为等比数列.
\(\because \dfrac{2^{a_{n+1}}}{2^{a_n}}=2^{a_{n+1}-a_n}\),不是常数,故② \(\{2^{a_n}\}\)不是等比数列.
\(\because \dfrac{\frac{1}{2 a_{n+1}}}{\frac{1}{2 a_n}}=\dfrac{a_n}{a_{n+1}}=\dfrac{1}{q},\),为常数,故③ \(\left\{\dfrac{1}{2 a_n}\right\}\)为等比数列,
故选:\(C\). -
证明 \(\because \lg a_n=3 n+5\), \(\therefore a_n=10^{3 n+5}\).
\(\therefore a_{n+1}=10^{3(n+1)+5}=10^{3 \mathrm{n}+8}\).
\(\therefore \dfrac{a_{n+1}}{a_n}=\dfrac{10^{3 n+8}}{10^{3 n+5}}=1000\).
\(\therefore\) 数列\(\{a_n \}\)是等比数列. -
证明 由\(2a_{n+1}=3a_n+1\),得\(a_{n+1}=\dfrac{3}{2} a_n+\dfrac{1}{2}\),即\(a_{n+1}+1=\dfrac{3}{2}\left(a_n+1\right)\),
故\(\dfrac{a_{n+1}+1}{a_n+1}=\dfrac{3}{2}\).
又\(a_1+1=2\),
所以\(\{a_n+1\}\)是首项为\(2\),公比为 \(\dfrac{3}{2}\)的等比数列.
【题型2】 等比数列的通项公式
【典题1】 在等比数列\(\{a_n \}\)中,\(a_5-a_1=15\),\(a_4-a_2=6\),则\(a_3=\)( )
A.\(-4\) \(\qquad \qquad \qquad \qquad\) B.\(4\) \(\qquad \qquad \qquad \qquad\) C.\(-4\)或\(4\) \(\qquad \qquad \qquad \qquad\) D.\(-8\)或\(8\)
解析 设等比数列的公比为\(q\),则
\(\because a_5-a_1=15\),\(a_4-a_2=6\),
\(\therefore\left\{\begin{array} { l }
{ a _ { 1 } q ^ { 4 } - a _ { 1 } = 1 5 } \\
{ a _ { 1 } q ^ { 3 } - a _ { 1 } q = 6 }
\end{array} \Rightarrow \left\{\begin{array} { c }
{ a _ { 1 } ( q ^ { 4 } - 1 ) = 1 5 } \\
{ a _ { 1 } ( q ^ { 3 } - q ) = 6 }
\end{array} \Rightarrow \left\{\begin{array}{c}
a_1\left(q^2-1\right)\left(q^2+1\right)=15 \\
a_1 q\left(q^2-1\right)=6
\end{array}\right.\right.\right.\),
\(\therefore q^2+1=\dfrac{5}{2} q\), \(\therefore q=2\)或 \(q=\dfrac{1}{2}\),
\(\therefore a_1=1\)或\(a_1=-16\),\(\therefore a_3=±4\),
故选:\(C\).
点拨 本题采取基本量法,\(a_1\),\(q\)是等比数列的基本量,在等比数列中遇到\(a_n\)采取通项公式 \(a_n=a_1q^{n-1}\);
由通项公式由已知条件得到关于\(a_1\),\(q\)的方程组,再通过因式分解消元解方程.
【典题2】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( )
A.\(-\dfrac{31}{32}\) \(\qquad \qquad \qquad \qquad\) B.\(-\dfrac{63}{64}\) \(\qquad \qquad \qquad \qquad\) C.\(-\dfrac{127}{128}\) \(\qquad \qquad \qquad \qquad\) D.\(-\dfrac{255}{256}\)
解析 设等比数列 \(\left\{\dfrac{1}{a_n+1}\right\}\)的公比为\(q\),\(a_1=7\),\(a_3=1\),
则 \(\dfrac{1}{1+1}=\dfrac{1}{1+7} \cdot q^2\),解得\(q^2=4\).
\(\therefore \dfrac{1}{a_{11}+1}=\dfrac{1}{a_1+1} \cdot q^{10}=\dfrac{1}{8} \times 2^{10}=2^7=128\),解得\(a_{11}=-\dfrac{127}{128}\),
故选:\(C\).
【巩固练习】
1.已知数列\(\{a_n \}\)满足\(a_1=1\),\(a_{n+1}=2a_n (n∈N^* )\),则\(a_4=\)( )
A.\(4\) \(\qquad \qquad \qquad \qquad\) B.\(6\) \(\qquad \qquad \qquad \qquad\) C. \(8\) \(\qquad \qquad \qquad \qquad\) D.\(2\)
2.设\(\{a_n \}\)是等比数列,下列结论中不正确的是( )
A.若\(a_1 a_2>0\),则\(a_2 a_3>0\) \(\qquad \qquad \qquad \qquad\) B.若\(a_1+a_3<0\),则\(a_5<0\) \(\qquad \qquad \qquad \qquad\)
C.若\(a_1 a_2<0\),则\(a_1 a_5<0\) \(\qquad \qquad \qquad \qquad\) D.若\(0<a_1<a_2\),则\(a_1+a_3>2a_2\)
3.已知\(-1\),\(a_1,\)\(a_2\),\(-4\)成等差数列,且\(-1\),\(b_1\),\(b_2\),\(b_3\),\(-4\)成等比数列,则 \(\dfrac{a_1+a_2}{b_2}\) 的值为\(\underline{\quad \quad}\).
4.\(\{a_n \}\)是各项均为正数的等差数列,\(\{b_n\}\)是等比数列,已知\(\dfrac{a_1}{b_1}=\dfrac{a_2}{b_2}=1\), \(\dfrac{a_3}{b_3}=\dfrac{8}{9}\),那么 \(\dfrac{a_4}{b_4}=\)\(\underline{\quad \quad}\).
参考答案
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答案 \(C\)
解析 因为\(a_1=1\),\(a_{n+1}=2a_n (n∈N^* )\),
所以 \(\dfrac{a_n+1}{a_n}=2\),即数列\(\{a_n \}\)是以\(1\)为首项,\(2\)为公比的等比数列,
所以\(a_n=2^{n-1}\),可得\(a_4=2^3=8\).
故选:\(C\). -
答案 \(C\)
解析 设等比数列\(\{a_n \}\)的公比为\(q\),
\(A\).\(\because a_1 a_2>0\),\(\therefore a_1^2 q>0\),\(\therefore q>0\),则\(a_2 a_3=a_1^2 q^3>0\),正确.
\(B\).\(\because a_1+a_3<0\),\(\therefore a_1 (1+q^2)<0\),\(\therefore a_1<0\),则\(a_5=a_1 q^4<0\),正确.
\(C\).\(\because a_1 a_2<0\),\(\therefore a_1^2 q<0\),\(\therefore q<0\),则\(a_1 a_5=a_1^2 q^4>0\),因此不正确.
\(D\).\(\because 0<a_1<a_2\),\(\therefore 0<a_1<a_1 q\),\(\therefore a_1>0\),\(q>0\),\(q≠1\).
则\(a_1+a_3=a_1 (1+q^2)>2a_1 q=2a_2\),正确.
故选:\(C\). -
答案 \(\dfrac{5}{2}\)
解析 以题意得\(a_1+a_2=-1+(-4)=-5\),\(b^2=-1×(-4)=4⇒b=±2\),
因\(-1\),\(b_1\),\(b_2\)成等比数列,\(\therefore b_1^2=-b_2>0⇒b_2<0\),
\(\therefore b_2=-2\),则 \(\dfrac{a_1+a_2}{b_2}=\dfrac{5}{2}\). -
答案 \(\dfrac{20}{27}\)
解析 设等差数列\(\{a_n \}\)的公差为\(d\),等比数列\(\{b_n\}\)的公比为\(q\),
则\(a_1+d=a_1 q\),\(9(a_1+2d)=8a_1 q^2\),
联立可得\(8q^2-18q+9=0\),解得: \(q=\dfrac{3}{2}\)或 \(q=\dfrac{3}{4}\).
\(\because \{a_n\}\)是各项均为正数,则\(d>0\),\(\therefore q>1\),则 \(q=\dfrac{3}{2}\),
\(\therefore b_4=a_1 \times\left(\dfrac{3}{2}\right)^3=\dfrac{27}{8} a_1\), \(\therefore \dfrac{a_2}{b_2}=\dfrac{3}{2} a_1\),
则\(d=\dfrac{3}{2} a_1-a_1=\dfrac{1}{2} a_1\),
\(\therefore a_4=a_1+3 d=\dfrac{5}{2} a\), \(\therefore \dfrac{a_4}{b_4}=\dfrac{\frac{5}{2} a_1}{\frac{27}{8} a_1}=\dfrac{20}{27}\).
【题型3】应用问题
【典题1】 某工厂\(2022\)年\(1\)月的生产总值为\(a\)万元,计划从\(2022\)年\(2\)月起,每个月生产总值比上一个月增长\(m\%\),那么到\(2023\)年\(8\)月底该厂的生产总值为多少万元?
解析 设从\(2022\)年\(1\)月开始,第\(n\)个月该厂的生产总值是\(a_n\)万元,则\(a_{n+1}=a_n+a_n m\%\),
\(\therefore \dfrac{a_{n+1}}{a_n}= 1+m\%\),
\(\therefore\)数列\(\{a_n \}\)是首项\(a_1=a\),公比\(q=1+m\%\)的等比数列.
\(\therefore a_n=a\left(1+m\%\right)^{n-1}\).
\(\therefore 2023\)年\(8\)月底该厂的生产总值为 \(a_{20}=a(1+m \%)^{20-1}=a(1+m \%)^{19}\) (万元).
【典题2】已知数列\(\{a_n \}\)满足\(S_n=n-a_n\).
(1)求证:数列\(\{a_n-1\}\)是等比数列;\(\qquad \qquad\) (2)求\(a_n\).
解析 (1)证明:\(\because\) 数列\(\{a_n \}\)满足\(S_n=n-a_n\).
\(\therefore S_{n+1}=n+1-a_{n+1}\),两式相减可得\(S_{n+1}-S_n={n+1}-n-a_{n+1}+a_n\),
\(\therefore a_{n+1}=1-a_{n+1}+a_n\), \(\therefore a_{n+1}=\dfrac{1}{2}+\dfrac{1}{2} a_n,\),
\(\therefore \dfrac{a_{n+1}-1}{a_n-1}=\dfrac{\dfrac{1}{2}+\dfrac{1}{2} a_n-1}{a_n-1}=\dfrac{\dfrac{1}{2}\left(a_n-1\right)}{a_n-1}=\dfrac{1}{2}\),
\(\therefore\) 数列\(\{a_n-1\}\)是\(\dfrac{1}{2}\)为公比的等比数列;
(2)由(1)可得数列\(\{a_n-1\}\)是\(\dfrac{1}{2}\)为公比的等比数列,
由\(S_n=n-a_n\)可得\(a_1=S_1=1-a_1\),解得\(a_1=\dfrac{1}{2}\),
故\(a_1-1=-\dfrac{1}{2}\),
\(\therefore a_n-1=\dfrac{1}{2}×\left(\dfrac{1}{2}\right)^{n-1}=\left(\dfrac{1}{2}\right)^n\),
\(\therefore a_n=1+\left(\dfrac{1}{2}\right)^n\).
【巩固练习】
1.河南洛阳龙门石窟是中国石刻艺术宝库,现为世界非物质文化遗产之一.某洞窟的浮雕共\(8\)层,它们构成一幅优美的图案.各层浮雕数成等比数列,第二层浮雕数为\(6\),第\(5\)层浮雕数为\(48\),则第\(7\)层浮雕数为( )
A.\(96\) \(\qquad \qquad \qquad \qquad\) B.\(128\) \(\qquad \qquad \qquad \qquad\) C. \(192\) \(\qquad \qquad \qquad \qquad\) D.\(384\)
2.我国古代数学著作《九章算术》中有“竹九节”问题:现有一根\(9\)节的竹子,自上而下各节的容积成等比数列,最上面\(3\)节的容积之积为\(3\)升,最下面\(3\)节的容积之积为\(243\)升,则第\(5\)节的容积是\(\underline{\quad \quad}\)升.
3.已知等比数列\(\{a_n \}\)的各项均为正数,且\(a_1>1\),前\(n\)项之积为\(T_n\),设 \(T_{10}=T_{20}\),
(1)当\(n\)为何值时,\(T_n\)最大?
(2)是否存在自然数\(n\),使得\(T_n=1\)?
4.在数列\(\{a_n \}\)中.已知\(a_1=2\), \(a_{n+1}=\dfrac{2 a_n}{a_n+1}\).
(1)求证: \(\left\{\dfrac{1}{a_n}-1\right\}\)是等比数列,
(2)若对任意\(n∈N_+\),\(a_n>m\)恒成立,求\(m\)的最大值.
参考答案
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答案 \(C\)
解析 设第\(2\)层浮雕数为\(a_2\),第\(5\)层浮雕数为\(a_5\),公比为\(q\),
则\(a_2=6\),\(a_5=48\),
\(\because\)各层浮雕数成等比数列,
\(\therefore 48=6×q^3\),\(\therefore q=2\),
\(\therefore\) 第\(7\)层浮雕数为\(a_7=a_5⋅q^2=48×4=192\),
故选:\(C\). -
答案 \(3\)
解析 设第\(n(n≤9,n∈N^* )\)节的容积为\(a_n\),则\(\{a_n \}\)是等比数列,
\(\because\)最上面\(3\)节的容积之积为\(3\)升,最下面\(3\)节的容积之积为\(243\)升,
\(\therefore\left\{\begin{array}{l} a_1 \cdot a_1 q^{\prime} a_1 q^2=3 \\ a_1 q^6 \cdot a_1 q^7 \cdot a_1 q^8=243 \end{array}\right.\),解得 \(a_1 q=3^{\frac{1}{3}}\), \(q^3=9^{\frac{1}{3}}\),
\(\therefore\)第\(5\)节的容积 \(a_5=a_1 q^4=a_1 q \cdot q^3=3^{\frac{1}{3}} \cdot 9^{\frac{1}{3}}=3\)(升). -
答案 (1) 当\(n=10\)时,\(T_n\)最大 ;(2) 当\(n=20\)时,\(T_{20}=1\).
解析 (1)设等比数列\(\{a_n \}\)的公比为\(q>0\),且\(a_1>1\),\(\because T_{10}=T_{20}\),
\(\therefore a_{11} a_{12} \cdot \ldots \cdot a_{20}=1\),
\(\therefore a_1^{10} q^{10+11+\cdots+19}=a_1^{10} q^{145}=1\),
\(\therefore a_1^2 q^{19}=1\).
\(\therefore T_n=a_1^n \cdot q^{1+2+\cdots+(n-1)}=a_1^n q^{\frac{n(n-1)}{2}}=a_1^{\frac{n(20-n)}{19}}\).
可知:当\(n=10\)时,指数\(\dfrac{n(20-n)}{19}\)取得最大值 \(\dfrac{100}{19}\),
\(\therefore\) 当\(n=10\)时,\(T_n\)最大.
(2)由 \(T_n=a_1^{\frac{n(20-n)}{19}}\),可知:当\(n=20\)时,\(T_{20}=1\). -
答案 (1)略 ;(2) \(1\) .
解析 证明:(1)\(\because a_1=2\), \(a_{n+1}=\dfrac{2 a_n}{a_n+1}\),
\(\therefore a_n>0\)恒成立;\(a_{n+1} a_n+a_{n+1}=2a_n\),
\(\therefore 1+\dfrac{1}{a_n}=2 \dfrac{1}{a_{n+1}}\) ,
\(\therefore \dfrac{1}{a_n}-1=2\left(\dfrac{1}{a_{n+1}}-1\right)\),
\(\therefore \dfrac{1}{a_{n+1}}-1=\dfrac{1}{2}\left(\dfrac{1}{a_n}-1\right)\),
又 \(\because \dfrac{1}{a_1}-1=-\dfrac{1}{2}\),
\(\therefore\left\{\dfrac{1}{a_n}-1\right\}\)是以\(-\dfrac{1}{2}\)为首项,\(\dfrac{1}{2}\)为公比的等比数列;
(2) \(\therefore\left\{\dfrac{1}{a_n}-1\right\}\)是以\(-\dfrac{1}{2}\)为首项,\(\dfrac{1}{2}\)为公比的等比数列,
\(\therefore \dfrac{1}{a_n}-1=-\dfrac{1}{2} \cdot\left(\dfrac{1}{2}\right)^{n-1}=-\left(\dfrac{1}{2}\right)^n\),
\(\therefore a_n=\dfrac{2^n}{2^n-1}=1+\dfrac{1}{2^n-1}\),
\(\therefore\)数列\(\{a_n \}\)是递减数列,且当\(n→+∞\)时,\(a_n→1\);
\(\therefore a_n>1\)恒成立,
\(\therefore m\)的最大值为\(1\).
分层练习
【A组---基础题】
1.在等比数列\(\{a_n \}\)中,已知\(a_1=\dfrac{9}{8}\),\(a_n=\dfrac{1}{3}\),\(q=\dfrac{2}{3}\),则\(n\)的值为( )
A.\(3\) \(\qquad \qquad \qquad \qquad\) B.\(4\) \(\qquad \qquad \qquad \qquad\) C. \(5\) \(\qquad \qquad \qquad \qquad\) D.\(6\)
2.在等比数列\(\{a_n \}\)中,已知\(a_1=2\),\(a_1-a_3+a_5=26\),则\(a_3=\)( )
A.\(20\) \(\qquad \qquad \qquad \qquad\) B.\(12\) \(\qquad \qquad \qquad \qquad\) C. \(8\) \(\qquad \qquad \qquad \qquad\) D.\(4\)
3.已知数列\(\{a_n \}\)是等比数列,下列四个命题中不正确的命题有( )
A.数列\(\{|a_n |\}\)是等比数列 \(\qquad \qquad \qquad \qquad\) B.数列\(\{a_n a_{n+1}\}\)是等比数列
C.数列 \(\left\{\dfrac{1}{a_n}\right\}\)是等比数列 \(\qquad \qquad \qquad \qquad\) D.数列\(\{\lg a_n^2\}\)是等比数列
4.标准对数视力表采用的“五分记录法”是我国独创的视力记录方式,此表由\(14\)行开口方向各异的正方形\(“E”\)形视标所组成,从上到下分别对应视力\(4.0,4.1,…,5.2,5.3\),且从第一行开始往下,每一行\(“E”\)形视标边长都是下一行\(“E”\)形视标边长的\(\sqrt[10]{10}\)倍,若视力\(4.1\)的视标边长为\(a\),则视力\(4.9\)的视标边长为( )
A. \(10^{\frac{2}{5}} a\) \(\qquad \qquad \qquad \qquad\) B. \(10^{-\frac{2}{5}} a\)\(\qquad \qquad \qquad \qquad\) C. \(10^{\frac{4}{5}} a\) \(\qquad \qquad \qquad \qquad\) D. \(10^{-\frac{4}{5}} a\)
5.(多选)已知等比数列\(\{a_n \}\)中,满足\(a_1=1\),公比\(q=-2\),则( )
A.数列\(\{2a_n+a_{n+1}\}\)是等比数列 \(\qquad \qquad \qquad \quad\) B.数列\(\{a_{n+1}-a_n\}\)是等比数列
C.数列\(\{a_n a_{n+1}\}\)是等比数列 \(\qquad \qquad \qquad \qquad\) D.数列\(\{ \log_2|a_n |\}\)是递减数列
6.一批设备价值\(a\)万元,由于使用磨损,每年比上一年价值降低\(b\%\),则\(n\)年后这批设备的价值为\(\underline{\quad \quad}\) .
7.设公差不为零的等差数列\(\{a_n \}\)满足\(a_3=7\),且\(a_1-1\),\(a_2-1\),\(a_4-1\)成等比数列,则\(a_{10}\)等于\(\underline{\quad \quad}\).
8.\(△ABC\),若\(\sin A\),\(\cos \dfrac{B}{2}\),\(\sin C\)成等比数列,则\(△ABC\)的形状为\(\underline{\quad \quad}\).
9.已知数列\(\{a_n \}\)中,\(a_1=1\),\(a_{n+1}=2a_n+1\),\((n∈N^*)\).求证:数列\(\{a_n+1\}\)是等比数列.
10.已知等比数列\(\{a_n \}\), \(a_1 a_2=-\dfrac{1}{2}\), \(a_3=\dfrac{1}{4}\).
(1)求数列\(\{a_n \}\)的通项公式;
(2)证明:对任意\(k∈N^*\),\(a_k\),\(a_{k+2}\), \(a_{k+1}\)成等差数列.
11.数列\(\{a_n \}\)满足 \(a_{n+1}=\dfrac{1}{2} a_n+1\),\(a_1=1\),若\(b_n=a_n-2\).
(1)求证:数列\(\{b_n \}\)是等比数列;\(\qquad \qquad\) (2)求数列\(\{a_n \}\)的通项公式.
参考答案
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答案 \(B\)
解析 \(\because\)在等比数列\(\{a_n \}\)中,已知 \(a_1=\dfrac{9}{8}\), \(a_n=\dfrac{1}{3}\), \(q=\dfrac{2}{3}\),
\(\therefore a_n=\dfrac{9}{8} \times\left(\dfrac{2}{3}\right)^{n-1}=\dfrac{1}{3}\), \(\therefore\left(\dfrac{2}{3}\right)^{n-1}=\dfrac{8}{27}=\left(\dfrac{2}{3}\right)^3\),
\(\therefore n-1=3\),可得\(n=4\),
故选:\(B\). -
答案 \(C\)
解析 根据题意,设等比数列\(\{a_n \}\)的公比为\(q\),
已知\(a_1=2\),\(a_1-a_3+a_5=26\),
则有\(a_1-a_3+a_5=a_1-a_1 q^2+a_1 q^4=2(1-q^2+q^4 )=26\),
解得\(q^2=4\)或\(-3\)(舍),
所以\(a_3=a_1 q^2=8\),
故选:\(C\). -
答案 \(D\)
解析 由\(\{a_n \}\)是等比数列可得 \(\dfrac{a_n}{a_{n-1}}=q\)\((q\)为常数,\(q≠0)\),
\(\dfrac{\left|a_n\right|}{\left|a_{n-1}\right|}=\left|\dfrac{a_n}{a_{n-1}}\right|=|q|\)为常数,故是等比数列;\(A\)正确.
\(\dfrac{a_n a_{n+1}}{a_{n-1} a_n}=\dfrac{a_{n+1}}{a_{n-1}}=q^2\)为常数,故是等比数列;\(B\)正确.
\(\dfrac{\frac{1}{a_n}}{\frac{1}{a_{n-1}}}=\dfrac{a_{n-1}}{a_n}=\dfrac{1}{q}\)常数,故是等比数列;\(C\)正确.
数列\(a_n=1\)是等比数列,但是\(\lg a_n^2=0\)不是等比数列;\(D\)不正确.
故选:\(D\). -
答案 \(D\)
解析 根据题意可知视际边长从上到下是以 \(\sqrt[10]{10}\)为公比的等比数列,
记视力\(4.1\)的视标边长为\(a_1=a\),
则视力\(4.9\)的视标边长为 \(a_9=a \cdot\left(10^{-\frac{1}{10}}\right)^8=10^{-\frac{4}{5}} a\).
故选:\(D\). -
答案 \(BC\)
解析 \(\because\) 等比数列\(\{a_n \}\)中,满足\(a_1=1\),公比\(q=-2\),
\(\therefore a_n=1×(-2)^{n-1}=(-2)^{n-1}\).
由此可得\(2a_n+a_{n+1}=2(-2)^{n-1}+(-2)^n=0\),故\(A\)错误;
\(a_{n+1}-a_n=(-2)^n-(-2)^{n-1}=-3(-2)^{n-1}\),
故数列\(\{a_{n+1}-a_n\}\)是等比数列,故\(B\)正确;
\(a_n a_{n+1}=(-2)^{n-1} (-2)^n=(-2)^{2n-1}\),故数列\(\{a_n a_{n+1}\}\)是等比数列,故\(C\)正确;
\(\log _2\left|a_n\right|=\log _2 2^{n-1}=n-1\),故数列\(\{ \log_2|a_n | \}\)是递增数列,故\(D\)错误,
故选:\(BC\). -
答案 \(a(1-b\%)^n\)
解析 依题意可知第一年后的价值为\(a(1-b\%)\),第二年价值为\(a(1-b\%)^2\),
依此类推可知每年的价值成等比数列,其首项\(a(1-b\%)\),公比为\(1-b\%\),
进而可知\(n\)年后这批设备的价值为\(a(1-b\%)^n\). -
答案 \(21\)
解析 设等差数列\(\{a_n \}\)的公差为\(d\),则\(d≠0\),
则\(a_1=a_3-2d=7-2d\),\(a_2=a_3-d=7-d\),\(a_4=a_3+d=7+d\),
由于\(a_1-1\),\(a_2-1\),\(a_4-1\)成等比数列,则\((a_2-1)^2=(a_1-1)(a_4-1)\),
即\((6-d)^2=(6-2d)(6+d)\),化简得\(d^2-2d=0\),由于\(d≠0\),解得\(d=2\),
因此\(a_{10}=a_3+7d=7+7×2=21\).
故答案为:\(21\). -
答案 等腰三角形
解析 \(\because \sin A\),\(\cos \dfrac{B}{2}\),\(\sin C\)成等比数列, \(\therefore \cos ^2 \dfrac{B}{2}=\sin A \sin C\),
\(\therefore \dfrac{1}{2}(1+\cos B)=-\dfrac{1}{2}[\cos (A+C)-\cos (A-C)]\),
\(\therefore 1+\cos B=-[-\cos B-\cos (A-C)]\),
化为 \(\cos (A-C)=1\),又\(A\),\(C∈(0,π)\),
\(\therefore A=C\),可得\(a=c\).
则\(△ABC\)的形状为等腰三角形. -
证明 \(\because a_{n+1}=2a_n+1\),\((n∈N^*)\),
\(\therefore a_{n+1}+1=2(a_n+1)\),
\(\therefore \dfrac{a_{n+1}+1}{a_n+1}=2\),
\(\therefore\)数列\(\{a_n+1\}\)是以\(2\)为公比的等比数列. -
答案 (1) \(a_n=\left(-\dfrac{1}{2}\right)^{n-1}\);(2) 略.
解析 (1)根据题意,设等比数列\(\{a_n \}\)的公比为\(q\),则\(a_n=a_1 q^{n-1}\),
若\(a_1 a_2=-\dfrac{1}{2}\),则 \(a_1^2 q=-\dfrac{1}{2}\),
若\(a_3=\dfrac{1}{4}\),则\(a_1 q^2=\dfrac{1}{4}\),变形可得\(\dfrac{a_1}{q}=-2\),解可得\(a_1^3=1\),则\(a_1=1\),
则有\(q=-\dfrac{1}{2}\) ;故\(a_n=\left(-\dfrac{1}{2}\right)^{n-1}\);
(2)证明:根据题意, \(a_n=\left(-\dfrac{1}{2}\right)^{n-1}\),
则\(a_k=\left(-\dfrac{1}{2}\right)^{k-1}\),\(a_{k+1}=\left(-\dfrac{1}{2}\right)^k\),\(a_{k+2}=\left(-\dfrac{1}{2}\right)^{k+1}\);
则\(a_k+a_{k+1}-2 a_{k+2}=\left(-\dfrac{1}{2}\right)^{k-1}+\left(-\dfrac{1}{2}\right)^k-2\left(-\dfrac{1}{2}\right)^{k+1}\)\(=\left(-\dfrac{1}{2}\right)^k\left[-2+1-2 x\left(-\dfrac{1}{2}\right)\right]=0\),
则有\(a_k+a_{k+1}=2 a_{k+2}\)
故\(a_k\),\(a_{k+2}\), \(a_{k+1}\)成等差数列. -
答案 (1)略 ;(2) \(a_n=2-\dfrac{1}{2^{n-1}}\).
解析 (1)证明: \(\because a_{n+1}=\dfrac{1}{2} a_n+1,\),
\(\therefore a_n-2=\dfrac{1}{2}\left(a_{n-1}-2\right)\),又\(b_n=a_n-2\),
\(\therefore b_n=\dfrac{1}{2} b_{n-1}\),
\(\therefore \{b_n \}\)是公式为\(\dfrac{1}{2}\)的等比数列;
(2)解:\(b_1=a_1-2=-1\), \(b_n=(-1) \times\left(\dfrac{1}{2}\right)^{n-1}\),
即\(a_n-2=-\left(\dfrac{1}{2}\right)^{n-1}\), \(\therefore a_n=2-\dfrac{1}{2^{n-1}}\) .
【B组---提高题】
1.已知实数\(a>0\),\(b>0\),\(\sqrt{2}\)是\(8^a\)与\(2^b\)的等比中项,则\(\dfrac{6}{a}+\dfrac{2}{b}\)的最小值是\(\underline{\quad \quad}\).
2.设数列\(\{a_n \}\)的首项\(a_1\)为常数,且\(a_n=3^{n-1}-2a_{n-1} (n≥2)\).
(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\(\{S_n\}\)是递增数列,求\(a_1\)的取值范围.
参考答案
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答案 \(32\)
解析 因为\(\sqrt{2}\)是\(8^a\)与\(2^b\)的等比中项,所以 \(8^a 2^b=(\sqrt{2})^2=2\),
即\(2^{3a+b}=2\),所以\(3a+b=1\),
所以\(\dfrac{6}{a}+\dfrac{2}{b}=\left(\dfrac{6}{a}+\dfrac{2}{b}\right)(3 a+b)=20+\dfrac{6 b}{a}+\dfrac{6 a}{b} \geq 20+2 \sqrt{\dfrac{6 b}{a} \cdot \dfrac{6 a}{b}}=32\),
当且仅当\(\dfrac{6 b}{a}=\dfrac{6 a}{b}\),即\(a^2=b^2\),\(a=b\)时,等号成立.
所以\(\dfrac{6}{a}+\dfrac{2}{b}\)的最小值是\(32\).
故答案为:\(32\). -
答案 (1) 略;(2) \(-\dfrac{3}{4}<a_1<\dfrac{3}{2}\) .
解析 (1)当\(n≥2\)时, \(a_n-\dfrac{3^n}{5}=3^{n-1}-2 a_{n-1}-\dfrac{3^n}{5}=2 \cdot \dfrac{3^{n-1}}{5}-2 a_{n-1}=-2\left(a_{n-1}-\dfrac{3^{n-1}}{5}\right)\)
(定义法证明等比数列,要注意首项\(a_1-\dfrac{3}{5}\)是否等于\(0\))
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\(-2\).
当\(a_1-\dfrac{3}{5}=0\),即 \(a_1=\dfrac{3}{5}\)时,数列\(\left\{a_n-\dfrac{3^n}{5}\right\}\)不是等比数列.
(2) \(a_1=\dfrac{3}{5}\)时, \(a_n=\dfrac{3^n}{5}\),为单调递增数列,满足条件.
\(a_1 \neq \dfrac{3}{5}\)时,由(1)可得: \(a_n-\dfrac{3^n}{5}=\left(a_1-\dfrac{3}{5}\right)(-2)^{n-1}\),
若\(\{S_n\}\)是递增数列,则\(S_n-S_{n-1}>0(n≥2)\),即\(a_n>0(n≥2\)),
\(\therefore a_n=\left(a_1-\dfrac{3}{5}\right)(-2)^{n-1}+\dfrac{3^n}{5}>0\),
(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 \(-2^n\left(a_1-\dfrac{3}{5}\right)+\dfrac{3^n}{5}>0 \Rightarrow a_1<\dfrac{2}{5} \cdot\left(\dfrac{3}{2}\right)^n+\dfrac{3}{5}\),
故\(a_1<\dfrac{2}{5} \cdot\left(\dfrac{3}{2}\right)^2+\dfrac{3}{5}=\dfrac{3}{2}\),( \(f(n)=\dfrac{2}{5} \cdot\left(\dfrac{3}{2}\right)^n+\dfrac{3}{5}\)是增数列, \(f_{\min }=f(2)\))
当\(n\)为奇数,则 \(2^n\left(a_1-\dfrac{3}{5}\right)+\dfrac{3^n}{5}>0 \Rightarrow a_1<-\dfrac{2}{5} \cdot\left(\dfrac{3}{2}\right)^n+\dfrac{3}{5}\),
故 \(a_1>-\dfrac{2}{5} \cdot\left(\dfrac{3}{2}\right)^3+\dfrac{3}{5}=-\dfrac{3}{4}\),( \(f(n)=-\dfrac{2}{5} \cdot\left(\dfrac{3}{2}\right)^n+\dfrac{3}{5}\)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: \(-\dfrac{3}{4}<a_1<\dfrac{3}{2}\).
【C组---拓展题】
1.数字\(1,2,3,…,n(n≥2)\)的任意一个排列记作\((a_1,a_2,…,a_n)\),设\(S_n\)为所有这样的排列构成的集合.集合\(A_n=\{(a_1,a_2,…,a_n)∈S_n |\)任意整数\(i\),\(j\),\(1≤i<j≤n\),都有\(a_i-i≤a_j-j\}\);集合\(B_n=\{(a_1,a_2,…,a_n\}∈S_n |\)任意整数\(i\),\(j\),\(1≤i<j≤n\),都有\(a_i+i≤a_j+j\}\).
(1)用列举法表示集合\(A_3\),\(B_3\)
(2)求集合\(A_n∩B_n\)的元素个数;
(3)记集合\(B_n\)的元素个数为\(b_n\).证明:数列\(\{b_n\}\)是等比数列.
参考答案
- 答案 (1)\(A_3=\{(1,2,3)\}\),\(B_3=\{(1,2,3),(1,3,2),(2,1,3),(3,2,1)\}\) ;
(2) \(1\) ;(3) 略.
解析 (1) (列出\(S_n\)所有项逐一检验,对于集合\(A_n\)可理解\(f(n)=a_n-n\)不是减数列,
对于集合\(B_n\)可理解\(f(n)=a_n+n\)不是减数列)
\(A_3=\{(1,2,3)\}\),\(B_3=\{(1,2,3),(1,3,2),(2,1,3),(3,2,1)\}\).
(2)考虑集合\(A_n\)中的元素\((a_1,a_2,…,a_n)\).
由已知,对任意整数\(1≤i<j≤n\),都有\(a_i-i≤a_j-j\),
所以\((a_i-i)+i<(a_j-j)+j\),
所以\(a_i<a_j\).
由\(i\),\(j\)的任意性可知,\((a_1,a_2,…,a_n)\)是\(1,2,3,…,n\)的单调递增排列,
所以\(A_n=\{(1,2,3,…,n)\}\).
又因为当\(a_k=k(k∈N^*,1≤k≤n)\)时,对任意整数\(i\),\(j\),\(1≤i<j≤n\),
都有\(a_i+i≤a_j+j\).
所以\((1,2,3,…,n)∈B_n\),所以\(A_n⊆B_n\).
所以集合\(A_n∩B_n\)的元素个数为\(1\).
(3)由(2)知,\(b_n≠0\).
因为\(B_2=\{(1,2),(2,1)\}\),所以\(b_2=2\).
当\(n≥3\)时,考虑\(B_n\)中的元素\((a_1,a_2,…,a_n)\).
(讨论\(n\)在\(B_n\)中的位置)
(1)假设\(a_k=n(1≤k<n)\).由已知,\(a_k+k≤a_{k+1}+(k+1)\),
所以\(a_{k+1}≥a_k+k-(k+1)=n-1\),
又因为\(a_{k+1}≤n-1\),所以\(a_{k+1}=n-1\).
依此类推,若\(a_k=n\),则\(a_{k+1}=n-1\),\(a_{k+2}=n-2\),…,\(a_n=k\).
①若\(k=1\),则满足条件的\(1,2,3,…,n\)的排列\((a_1,a_2,…,a_n)\)有\(1\)个.
②若\(k=2\),则\(a_2=n\),\(a_3=n-1\),\(a_4=n-2\),…,\(a_n=2\).
所以\(a_1=1\).
此时满足条件的\(1,2,3,…,n\)的排列\((a_1,a_2,…,a_n)\)有\(1\)个.
③若\(2<k<n\),
只要\((a_1,a_2,a_3,…a_{k-1})\)是\(1,2,3,…,k-1\)的满足条件的一个排列,就可以相应得到\(1,2,3,…,n\)的一个满足条件的排列.
(把问题转化为\(n=k-1\)的情况,这方法巧妙,可得到\(b_n\)的一递推公式)
此时,满足条件的\(1,2,3,…,n\)的排列\((a_1,a_2,…,a_n)\)有\(b_{k-1}\)个.
(2)假设\(a_n=n\),只需\((a_1,a_2,a_3,…a_{n-1})\)是\(1,2,3,…,n-1\)的满足条件的排列,此时满足条件的\(1,2,3,…,n\)的排列\((a_1,a_2,…,a_n)\)有\(b_{n-1}\)个.
综上\(b_n=1+1+b_2+b_3+⋯+b_{n-1}\),\(n≥3\).
因为\(b_3=1+1+b_2=4=2b_2\),
且当\(n≥4\)时,\(b_n=(1+1+b_2+b_3+⋯+b_{n-2})+ b_{n-1}=2b_{n-1}\),
所以对任意\(n∈N^*\),\(n≥3\),都有 \(\dfrac{b_n}{b_{n-1}}=2\).
所以\(\{b_n\}\)成等比数列.