4.3.2 等比数列的前n项和公式
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基础知识
等比数列的前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
(1) 证明 等比数列\(\{a_n \}\)的首项为\(a_1\),公比为\(q\),则其前\(n\)项和是
\(S_n=a_1+a_2+a_3+⋯+a_n=a_1+a_1 q+a_1 q^2+⋯+a_1 q^{n-1}\) (1),
两边乘以公比\(q\)得\(qS_n=a_1 q+a_1 q^2+⋯+a_1 q^{n-1} +a_1 q^n\) (2)
(1)-(2)得\((1-q)S_n=a_1 (1-q^n )\),
当\(q≠1\)时, \(S_n=\dfrac{a_1\left(1-q^n\right)}{1-q}\);
当\(q=1\)时,\(S_n=a_1+a_2+a_3+⋯+a_n=a_1+a_1+a_1+⋯+a_1=na_1\),
故等比数列的前\(n\)项和为 \(S_n= \begin{cases}n a_1 & (q=1) \\ \dfrac{a_1\left(1-q^n\right)}{1-q} & (q \neq 1)\end{cases}\).
以上的方法称之为 错位相减法.
(2)当公比\(q=1\)时,\(S_n=na_1\),是\(n\)的正比例函数;
当公比\(q≠1\)时,\(S_n=\dfrac{a_1\left(1-q^n\right)}{1-q}\),它可变形为\(\ S_n=-\dfrac{a_1}{1-q} \cdot q^n+\dfrac{a_1}{1-q}\),
设 \(A=\dfrac{a_1}{1-q}\),上式可写成\(S_n=-Aq^n+A\).
由此可见,非常数列的等比数列的前\(n\)项和\(S_n\)是由关于\(n\)的一个指数式与一个常数的和构成的,
而指数式的系数与常数项互为相反数.
即若某数列的前\(n\)项和公式为\(S_n=-A⋅q^n+A\) \((A≠0,q≠0\)且\(q≠1,n∈N^*)\),则此数列一定是等比数列.
【例】等比数列\(\{a_n \}\)的公比\(q=2\),首项\(a_1=2\),则\(S_n\)等于\(\underline{\quad \quad}\) .
答案 \(2^{n+1} -2\)
基本性质
(1)若\(q≠-1\),则\(S_n\) , \(S_{2 n}-S_n\) , \(S_{3 n}-S_{2 n}\),…成等比数列,且公比\(q^n\);
\((q=-1\),\(n\)是偶数时,\(S_n=0)\)
证明 \(\dfrac{S_{(k+1) n}-S_{k n}}{S_{k n}-S_{(k-1) n}}=\dfrac{a_{k n+1}+a_{k n+2}+\cdots+a_{(k+1) n}}{a_{(k-1) n+1}+a_{(k-1) n+2}+\cdots+a_{k n}}\)\(=\dfrac{q^n\left(a_{(k-1) n+1}+a_{(k-1) n+2}+\cdots+a_{k n}\right)}{a_{(k-1) n+1}+a_{(k-1) n+2}+\cdots+a_{k n}}=q^n\).
(2)在等比数列\(\{a_n \}\)中,当总项数为\(2n\)时, \(S_{\text {偶 }}=q S_{\text {奇 }}\).
证明 \(S_{\text {偶 }}=a_2+a_4+\cdots+a_{2 n}=q\left(a_1+a_3+\cdots+a_{2 n-1}\right)=q S_{\text {奇 }}\).
(3) \(S_{n+m}=S_n+q^n S_m\).
证明 \(S_{m+n}=a_1+a_2+\cdots+a_n+a_{n+1}+\cdots+a_{n+m-1}+a_{n+m}\)
\(=S_n+a_1⋅q^n+a_2⋅q^n+a_3⋅q^n+⋯+a_m⋅q^n\)
\(=S_n+q^n (a_1+a_2+⋯+a_m )\)
\(=S_n+q^n S_m\).
基本方法
【题型1】 等比数列前n项和的有关计算问题
【典题1】 记正项等比数列\(\{a_n \}\)的前\(n\)项和是\(S_n\),若\(4(a_1+a_2)=3\), \(S_6=\dfrac{63}{4}\),则\(a_7=\)( )
A.\(32\) \(\qquad \qquad \qquad \qquad\) B.\(16\) \(\qquad \qquad \qquad \qquad\) C. \(\dfrac{1}{128}\) \(\qquad \qquad \qquad \qquad\) D.\(\dfrac{1}{256}\)
解析 根据题意,显然等比数列\(\{a_n \}\)的公比\(q≠1\),故 \(S_6=\dfrac{a_1\left(1-q^6\right)}{1-q}=\dfrac{a_1-a_7}{1-q}=\dfrac{63}{4}\) ①,
又知道\(4\left(a_1+a_2\right)=4 S_2=\dfrac{4 a_1\left(1-q^2\right)}{1-q}=3\) ②,
①÷②得:\(q^4+q^2-20=0\),解得\(q^2=4\)或\(q^2=-5\)(舍),
因为数列\(\{a_n \}\)是正项等比数列,所以\(q=2\),
由\(4(a_1+a_2)=4(a_1+2a_1)=3\),得 \(a_1=\dfrac{1}{4}\),
所以 \(a_7=a_1 \cdot q^6=\dfrac{1}{4} \times 2^6=16\).
故选:\(B\).
【典题2】 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( )
A.\([-1,0]\) \(\qquad \qquad \qquad\) B.\(\left[-1,\dfrac{1}{2}\right]\) \(\qquad \qquad \qquad\) C.\(\left[\dfrac{1}{2},1\right]\) \(\qquad \qquad \qquad\) D.\([0,1]\)
解析 设等比数列\(\{a_n \}\)的公比为\(q\),\(\because a_1+2a_2=0\),\(S_3=\dfrac{3}{4}\),
\(\therefore a_1 (1+2q)=0\), \(a_1\left(1+q+q^2\right)=\dfrac{3}{4}\),解得:\(a_1=1\),\(q=-\dfrac{1}{2}\),
\(\therefore S_n=\dfrac{1-\left(-\dfrac{1}{2}\right)^n}{1-\left(-\dfrac{1}{2}\right)}=\dfrac{2}{3}\left[1-\left(-\dfrac{1}{2}\right)^n\right]\).
当\(n=1\)时,\(S_n\)取最大值\(1\),当\(n=2\)时,\(S_n\)取最小值\(\dfrac{1}{2}\),
\(\therefore\left\{\begin{array}{l}
a \leq \dfrac{1}{2} \\
a+2 \geq 1
\end{array}\right.\),\(-1≤a≤\dfrac{1}{2}\),
故选:\(B\).
【典题3】 已知等比数列\(\{a_n \}\)的公比为\(2\),若\(a_1+a_3=5\),则\(a_1 a_2+a_2 a_3+a_3 a_4+⋯+a_{99} a_{100}=\)\(\underline{\quad \quad}\).
解析 因为等比数列\(\{a_n \}\)的公比为\(2\),\(a_1+a_3=5\),
所以\(a_1 (1+q^2 )=5\),可得\(a_1=1\),则 \(a_n=2^{n-1}\) , \(a_n a_{n+1}=2^{2 n-1}\),
则 \(a_1 a_2+a_2 a_3+a_3 a_4+\cdots+a_{99} a_{100}=\dfrac{2}{3}\left(4^{99}-1\right)\).
【巩固练习】
1.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\(S_{10}=\)( )
A.\(16\) \(\qquad \qquad \qquad \qquad\) B.\(32\) \(\qquad \qquad \qquad \qquad\) C.\(40\) \(\qquad \qquad \qquad \qquad\) D.\(62\)
2.等比数列\(\{a_n \}\)的前\(n\)项和为\(S_n\),\(a_1=a_2+2a_3\),\(S_2\)是\(S_1\)与\(mS_3\)的等比中项,则\(m\)的值为( )
A.\(1\) \(\qquad \qquad \qquad \qquad\) B.\(\dfrac{9}{7}\) \(\qquad \qquad \qquad \qquad\) C. \(\dfrac{6}{7}\) \(\qquad \qquad \qquad \qquad\) D.\(\dfrac{1}{2}\)
3.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\(a_8=2\),则\(a_2+a_5=\)( )
A.\(-12\) \(\qquad \qquad \qquad \qquad\) B.\(-4\) \(\qquad \qquad \qquad \qquad\) C.\(4\) \(\qquad \qquad \qquad \qquad\) D.\(12\)
4.已知等比数列\(\{a_n \}\)的前\(n\)项和是\(S_n\),若\(3a_1+2a_2=4\),\(9S_3=8S_6\),则\(S_5=\)( )
A.\(\dfrac{15}{8}\)或\(5\) \(\qquad \qquad \qquad \qquad\) B.\(\dfrac{31}{16}\)或\(5\) \(\qquad \qquad \qquad \qquad\) C. \(\dfrac{31}{16}\) \(\qquad \qquad \qquad \qquad\) D.\(\dfrac{15}{8}\)
5.已知等比数列\(\{a_n \}\)的前\(n\)项和为\(S_n\),则下列结论一定成立的是( )
A.若\(a_5>0\),则\(a_{2017}<0\) \(\qquad \qquad \qquad \qquad\) B.若\(a_6>0\),则\(a_{2018}<0\)
C.若\(a_5>0\),则\(S_{2017}>0\) \(\qquad \qquad \qquad \qquad\) D.若\(a_6>0\),则\(S_{2018}>0\)
参考答案
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答案 \(D\)
解析 设公比为\(q\),由\(a_1+a_2=2,a_5+a_6=8\), \(\left\{\begin{array}{l} a_1+a_1 q=2 \\ a_1 q^4+a_1 q^5=8 \end{array}\right.\),
解得 \(a_1=2 \sqrt{2}-2\), \(q=\sqrt{2}\),或 \(q=-\sqrt{2}\)(舍去),
\(S_{10}=\dfrac{(2 \sqrt{2}-2)\left(1-(\sqrt{2})^{10}\right)}{1-\sqrt{2}}=62\),
故选:\(D\). -
答案 \(B\)
解析 设数列\(\{a_n \}\)的公比为\(q\),则由\(a_1=a_2+2a_3\),得\(a_1=a_1 q+2a_1 q^2\),
易知\(a_1≠0\),所以\(2q^2+q-1=0\),解得\(q=-1\)或\(q=\dfrac{1}{2}\),
当\(q=-1\)时,\(S_2=0\),这与\(S_2\)是\(S_1\)与\(mS_3\)的等比中项矛盾
当\(q=\dfrac{1}{2}\)时,\(S_1=a_1\), \(S_2=\dfrac{3}{2} a_1\), \(m S_3=\dfrac{7}{4} a_1 m\),
由\(S_2\)是\(S_1\)与\(mS_3\)的等比中项,得\(S_2^2=S_1⋅mS_3\),
即\(\dfrac{9}{4} a_1^2=m \cdot \dfrac{7}{4} a_1^2\),所以 \(m=\dfrac{9}{7}\),
故选:\(B\). -
答案 \(C\)
解析 由题意可得:等比数列\(\{a_n\}\)的\(q≠1\),
\(\because S_3\),\(S_9\),\(S_6\)成等差数列,且\(a_8=2\),
\(\therefore 2S_9=S_6+S_3\),且\(a_8=2\),
\(\therefore 2 \times \dfrac{a_1\left(q^9-1\right)}{q-1}=\dfrac{a_1\left(q^6-1\right)}{q-1}+\dfrac{a_1\left(q^3-1\right)}{q-1}\),\(a_1 q^7=2\).
解得:\(q^3=-\dfrac{1}{2}\),\(a_1 q=8\).
则\(a_2+a_5=a_1 q(1+q^3)=8×(1-\dfrac{1}{2})=4\).
故选:\(C\). -
答案 \(C\)
解析 等比数列\(\{a_n \}\)的前\(n\)项和是\(S_n\),\(3a_1+2a_2=4\),\(9S_3=8S_6\),
\(\therefore\) 当公比\(q=1\)时,\(\left\{\begin{array}{l} 3 a_1+2 a_1=4 \\ 9 \times 3 a_1=8 \times 6 a_1 \end{array}\right.\),无解,
当公比\(q≠1\)时, \(\left\{\begin{array}{l} 3 a_1+2 a_1 q=4 \\ 9 \times \dfrac{a_1\left(1-q^3\right)}{1-q}=8 \times \dfrac{a_1\left(1-q^6\right)}{1-q} \end{array}\right.\),解得\(a_1=1\),\(q=\dfrac{1}{2}\),
\(\therefore S_5=\dfrac{1 \times\left(1-\dfrac{1}{2^5}\right)}{1-\dfrac{1}{2}}=\dfrac{31}{16}\).
故选:\(C\). -
答案 \(C\)
解析 \(A\).若\(a_5=a_1 q^4>0\),则 \(a_{2017}=a_1 q^{2016}>0\),故本选项错误;
\(B\)、\(a_6=a_1 q^5>0\),则 \(a_{2018}=a_1 q^{2017}>0\),故本选项错误;
\(C\)、\(a_5=a_1 q^4>0\),则\(a_1>0\).若\(q=1\)时, \(S_{2017}=2017 a_1>0\);
若\(q≠1\), \(S_{2017}=\dfrac{a_1\left(q^{2017}-1\right)}{q-1}>0\),故本选项正确;
\(D\)、若\(a_6=a_1 q^5>0\).若\(a_1>0\),\(q=1\)时, \(S_{2018}=2018 a_1>0\);
若\(q≠1\), \(S_{2018}=\dfrac{a_1\left(q^{2018}-1\right)}{q-1}\),
当\(q<-1\)时, \(S_{2018}<0\),\(q=-1\), \(S_{2018}=0\),故本选项错误;
故选:\(C\).
【题型2】 等比数列前n项和的性质应用
【典题1】 在等比数列\(\{a_n \}\)中,已知\(S_n=48\),\(S_{2n}=60\),求\(S_{3n}.\)
解析 方法1 \(\because S_{2n}≠2S_n\), \(\therefore q≠1\).
由已知,得\(\left\{\begin{array}{l}
\dfrac{a_1\left(1-q^n\right)}{1-q}=48 \\
\dfrac{a_1\left(1-q^{2 n}\right)}{1-q}=60
\end{array}\right.\),
\(\therefore 1+q^n=\dfrac{5}{4}\),即 \(q^n=\dfrac{1}{4}\),代入 \(\dfrac{a_1\left(1-q^n\right)}{1-q}=48\)得 \(\dfrac{a_1}{1-q}=64\).
故\(S_{3 n}=\dfrac{a_1\left(1-q^{3 n}\right)}{1-q}=64 \times\left(1-\dfrac{1}{4^3}\right)=63\).
方法2 \(\because \{a_n \}\)为等比数列,
\(\therefore S_n\),\(S_{2n}-S_n\),\(S_{3n}-S_{2n}\)也成等比数列,
\(\therefore (S_{2n}-S_n )^2=S_n (S_{3n}-S_{2n} )\),
\(\therefore S_{3 n}=\dfrac{\left(S_{2 n}-S_n\right)^2}{S_n}+S_{2 n}=\dfrac{(60-48)^2}{48}+60=63\).
点拨 方法2采取等比数列前\(n\)项和的性质:\(S_n\),\(S_{2n}-S_n\),\(S_{3n}-S_{2n}\),\(…\)成等比数列,较方法1简便.
【巩固练习】
1.设等比函数\(\{a_n \}\)的前\(n\)项和为\(S_n\),若\(\dfrac{S_6}{S_3}=3\),则\(\dfrac{S_{12}}{S_9}=\)( )
A. \(\dfrac{7}{3}\) \(\qquad \qquad \qquad \qquad\) B. \(\dfrac{15}{7}\)\(\qquad \qquad \qquad \qquad\) C. \(\dfrac{15}{7}\) \(\qquad \qquad \qquad \qquad\) D. \(\dfrac{8}{3}\)
2.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( )
A.\(90\) \(\qquad \qquad \qquad \qquad\) B.\(60\) \(\qquad \qquad \qquad \qquad\) C.\(45\) \(\qquad \qquad \qquad \qquad\) D.\(32\)
参考答案
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答案 \(B\)
解析 \(\because \{a_n\}\)为等比数列,则\(S_3\),\(S_6-S_3\),\(S_9-S_6\),\(S_{12}-S_9\)也成等比数列,
由\(S_6:S_3=1:3\),
令\(S_3=x\),则 \(S_6=\dfrac{1}{3} x\),
则由\(x\), \(-\dfrac{2}{3} x\), \(S_9-\dfrac{1}{3} x\), \(S_{12}-S_9\)也成等比数列,
可得 \(S_9=\dfrac{7}{9} x\),\(S_{12}=\dfrac{5}{3} x\),则 \(\dfrac{S_{12}}{S_9}=\dfrac{15}{7}\).
故选:\(B\). -
答案 \(C\)
解析 \(\because\)等比数列\(\{a_n \}\)中,\(a_1+a_2+a_3=3\),\(a_4+a_5+a_6=6\),
由等比数列的性质得:
\(a_1+a_2+a_3\),\(a_4+a_5+a_6\),\(a_7+a_8+a_9\),\(a_{10}+a_{11}+a_{12}\)也成等比数列,
\(\therefore\)由\(a_1+a_2+a_3=3\),\(a_4+a_5+a_6=6\),得\(a_7+a_8+a_9=12\), \(a_{10}+a_{11}+a_{12}=24\),
\(\therefore \{a_n\}\)的前\(12\)项和为:\(3+6+12+24=45\).
故选:\(C\).
【题型3】 综合问题
【典题1】 已知等比数列\(\{a_n \}\)的前\(n\)项和为\(S_n\),\(a_1=-1\), \(\dfrac{S_{10}}{S_5}=\dfrac{31}{32}\).
(1)求等比数列\(\{a_n \}\)的公比\(q\); \(\qquad \qquad\) (2)求\(a_1^2+a_2^2+⋯+a_n^2\).
解析 (1)由题意可知\(q≠1\),
因为\(\dfrac{S_{10}}{S_5}=\dfrac{31}{32}\),所以\(\dfrac{\dfrac{a_1\left(1-q^{10}\right)}{1-q}}{\dfrac{a_1\left(1-q^5\right)}{1-q}}=1-q^5=\dfrac{31}{32}\),解得\(q=\dfrac{1}{2}\).
(2)由(1)可得\(a_n=-\dfrac{1}{2^{n-1}}\),则 \(a_n^2=\dfrac{1}{4^{n-1}}\) , \(\dfrac{a_n^2}{a_{n-1}^2}=\dfrac{1}{4}\),
所以数列\(\{a_n^2 \}\)是首项为\(1\),公比为\(\dfrac{1}{4}\)的等比数列,
所以\(a_1^2+a_2^2+\cdots+a_n^2=\dfrac{1-\dfrac{1}{n}}{1-\dfrac{1}{4}}=\dfrac{4}{3}-\dfrac{1}{3 \cdot 4^{n-1}}\).
【典题2】 某地本年度旅游业收入估计为\(400\)万元,由于该地出台了一系列措施,进一步发展旅游业,预计今后旅游业的收入每年会比上一年增加\(\dfrac{1}{4}\).
(1)求\(n\)年内旅游业的总收入;
(2)试估计大约几年后,旅游业的总收入超过\(8 000\)万元.
解析 设第\(n\)年的旅游业收入估计为\(a_n\)万元,
则\(a_1=400\), \(a_{n+1}=\left(1+\dfrac{1}{4}\right) a_n=\dfrac{5}{4} a_n\),
\(\therefore \dfrac{a_{n+1}}{a_n}=\dfrac{5}{4}\), \(\therefore \{a_n\}\)是公比为\(\dfrac{5}{4}\)的等比数列.
\(\therefore S_n=\dfrac{a_1\left(1-q^n\right)}{1-q}=\dfrac{400\left[1-\left(\dfrac{5}{4}\right)^n\right]}{1-\dfrac{5}{4}}=1600\left[\left(\dfrac{5}{4}\right)^n-1\right]\),
即\(n\)年内旅游业总收入为\(1600\left[\left(\dfrac{5}{4}\right)^n-1\right]\)万元.
(2)由(1)知\(S_n=1600\left[\left(\dfrac{5}{4}\right)^n-1\right]\),令\(S_n>8000\),
即\(1600\left[\left(\dfrac{5}{4}\right)^n-1\right]>8000\),
\(\therefore\left(\dfrac{5}{4}\right)^n>6\), \(\therefore \lg \left(\dfrac{5}{4}\right)^n>\lg 6\).
\(\therefore n>\dfrac{\lg 6}{\lg \dfrac{5}{4}}\approx 8.0296\).
\(\therefore\)大约第\(9\)年后,旅游业总收入超过\(8 000\)万元.
【巩固练习】
1.中国古代数学著作《算法统宗》中,载有一些关于等比数列的问题.如:“三百七十八里关,初行健步不为难,次日脚痛减一半,六朝才得到其关……”问:此人最后一天走的路程是( )
A .\(4\)里 \(\qquad \qquad \qquad \qquad\) B.\(5\)里 \(\qquad \qquad \qquad \qquad\) C. \(6\)里 \(\qquad \qquad \qquad \qquad\) D.\(8\)里
2.我国古代数学著作《九章算术》中有如下问题:“今有女子善织,日自倍,五日织五尺,问日织几何?”意思是:“一女子善于织布,每天织出的布都是前一天的\(2\)倍,已知她\(5\)天共织布\(5\)尺,问这女子每天织布多少?”这个问题体现了古代对数列问题的研究.某数学爱好者对于这道题作了以下改编:有甲、乙两位女子,需要合作织出\(40\)尺布.两人第一天都织出一尺,以后几天中,甲女子每天织出的布都是前一天的\(2\)倍,乙女子每天织出的布都比前一天多半尺,则两人完成织布任务至少需要( )
A.\(2\) 天 \(\qquad \qquad \qquad \qquad\) B.\(3\)天 \(\qquad \qquad \qquad \qquad\) C. \(4\)天 \(\qquad \qquad \qquad \qquad\) D.\(5\)天
3.如图,在边长为\(1\)的等边\(△ABC\)中,圆\(O_1\)为\(△ABC\)的内切圆,圆\(O_2\)与圆\(O_1\)外切,且与\(AB\),\(BC\)相切,…,圆\(O_{n+1}\)与圆\(O_n\)外切,且与\(AB\),\(BC\)相切,如此无限继续下去.记圆\(O_n\)的面积为\(a_n (n∈N)\).
(1)证明\(\{a_n \}\)是等比数列; \(\qquad \qquad\)(2)求\(a_1+a_2+⋯+a_n\)的值.
4.已知单调递减的等比数列\(\{a_n \}\)的前\(n\)项和为\(S_n\), \(a_1+a_3=\dfrac{5}{8}\),\(S_3=7a_3\).
(1)求数列\(\{a_n \}\)的通项公式;
(2)求满足 \(S_n \leq \dfrac{999}{1000}\)的所有正整数\(n\)的值.
5.已知公比小于\(1\)的等比数列\(\{a_n \}\)中,其前\(n\)项和为\(S_n\),\(a_2=\dfrac{1}{4}\), \(S_3=\dfrac{7}{8}\).
(1)求\(a_n\);\(\qquad \qquad\) (2)求证:\(\dfrac{1}{2}≤S_n<1\).
参考答案
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答案 \(C\)
解析 由题意得该人第\(n\)天行走的路程\(\{a_n \}\)构成公比为\(q=\dfrac{1}{2}\)的等比数列,
且这个等比数列的前\(6\)项和\(S_6=378\),
\(\therefore S_6=\dfrac{a_1\left(1-\dfrac{1}{6}\right)}{1-\dfrac{1}{2}}=378\),解得\(a_1=192\),
\(\therefore\) 此人最后一天走的路程是\(a_6=192×(\dfrac{1}{2})^5=6\)(里).
故选:\(C\). -
答案 \(D\)
解析 设甲,乙每天织布分别记为数列\(\{a_n \}\),\(\{b_n \}\),
由题意得数列\(\{a_n \}\)是以\(1\)为首项,\(2\)为公比的等比数列,
\(\{b_n \}\)是以\(1\)为首项,以\(\dfrac{1}{2}\)为公差的等差数列,
故\(n+\dfrac{1}{4} n(n-1)+\dfrac{1-2^n}{1-2} \geq 40\),
即\(n^2+3n+2^{n+2}≥164\),
经检验\(n=5\)时符合题意.
故选:\(D\). -
答案 (1)略;(2) \(a_1+a_2+\cdots+a_n=\dfrac{\sqrt{3}}{4}\left[1-\left(\dfrac{1}{3}\right)^n\right]\).
解析 (1)证明:记\(a_1\)为圆\(O_n\)的半径,则 \(a_1=\dfrac{1}{2} \tan 30^{\circ}=\dfrac{\sqrt{3}}{6},\),
又 \(\dfrac{a_{n-1}-a_n}{a_{n-1}+a_n}=\sin 30^{\circ}=\dfrac{1}{2} \text {, }\), \(\therefore a_n=\dfrac{1}{3} a_{n-1} (n≥2)\),
所以\(\{a_n \}\)是首项为 \(\dfrac{\sqrt{3}}{6}\),公比为\(\dfrac{1}{3}\)的等比数列;
(2)由(1)得 \(a_n=\dfrac{\sqrt{3}}{6} \times\left(\dfrac{1}{3}\right)^{n-1}=\dfrac{3^{\dfrac{1}{2}-n}}{2}\),
所以\(a_1+a_2+\cdots+a_n=\dfrac{\dfrac{\sqrt{3}}{6}\left[1-\left(\dfrac{1}{3}\right)^n\right]}{1-\dfrac{1}{3}}=\dfrac{\sqrt{3}}{4}\left[1-\left(\dfrac{1}{3}\right)^n\right]\).
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答案 (1) \(a_n=\dfrac{1}{2^n}\) ;(2) \(1、2、3、4、5、6、7、8、9\).
解析 (1)根据题意,设等比数列\(\{a_n \}\)的公比为\(q\),\((0<q<1)\),
又由\(a_1+a_3=\dfrac{5}{8}\),\(S_3=7a_3\),则有\(\left\{\begin{array}{l} a_1+a_1 q^2=\dfrac{5}{8} \\ a_1+a_1 q+a_1 q^2=7 \times a_1 q^2 \end{array}\right.\),
解可得\(a_1=\dfrac{1}{2}\),\(q=\dfrac{1}{2}\),
故\(a_n=a_1 q^{n-1} =\dfrac{1}{2^n}\) ;
(2)根据题意,由(1)的结论,\(a_n=\dfrac{1}{2^n}\) ,则\(S_n=\dfrac{a_1\left(1-q^n\right)}{1-q}=1-\dfrac{1}{2^n}\),
若\(S_n \leq \dfrac{999}{1000}\),即\(1-\dfrac{1}{2^n} \leq \dfrac{999}{1000}\),变形可得\(2^n≤1000\),
又由\(n∈N^+\),则\(n≤9\),
即\(n\)可取的值为\(1、2、3、4、5、6、7、8、9\). -
答案 (1) \(a_n=\left(\dfrac{1}{2}\right)^n\);(2) 略.
解析 (1)解:设等比数列\(\{a_n \}\)的公比为\(q\).
由 \(\left\{\begin{array}{l} a_2=\dfrac{1}{4} \\ S_3=\dfrac{7}{8} \end{array}\right.\)或 \(\left\{\begin{array}{l} a_2=\dfrac{1}{4} \\ \dfrac{1}{4 q}+\dfrac{1}{4}+\dfrac{1}{4} q=\dfrac{7}{8} \end{array}\right.\),
解得 \(\left\{\begin{array}{l} a_2=\dfrac{1}{4} \\ q=\dfrac{1}{2} \end{array}\right.\)或 \(\left\{\begin{array}{l} a_2=\dfrac{1}{4} \\ q=2 \end{array}\right.\)(舍去),
所以\(a_n=\dfrac{1}{4}×\left(\dfrac{1}{2}\right)^{n-2} =\left(\dfrac{1}{2}\right)^n\).
(2)证明:由(1)得\(a_n=\left(\dfrac{1}{2}\right)^n\),
所以 \(S_n=\dfrac{\dfrac{1}{2}\left[1-\left(\dfrac{1}{2}\right)^n\right]}{1-\dfrac{1}{2}}=1-\left(\dfrac{1}{2}\right)^n\).
因为\(y=\left(\dfrac{1}{2}\right)^x\)在\(R\)上为减函数,且\(y=\left(\dfrac{1}{2}\right)^x>0\)恒成立,
所以当\(n∈N^*\),即\(n≥1\)时,\(0<\left(\dfrac{1}{2}\right)^n≤\dfrac{1}{2}\),
所以\(\dfrac{1}{2}≤S_n=1-\left(\dfrac{1}{2}\right)^n<1\).
分层练习
【A组---基础题】
1.已知等比数列\(\{a_n \}\)的公比\(q>0\),前\(n\)项和是\(S_n\),若\(a_1=1\),\(S_4=5S_2\),则\(a_5=\)( )
A.\(16\) \(\qquad \qquad \qquad \qquad\) B.\(12\) \(\qquad \qquad \qquad \qquad\) C.\(8\) \(\qquad \qquad \qquad \qquad\) D.\(4\)
2.已知\(S_n\)是等比数列\(\{a_n \}\)的前\(n\)项和,若\(S_5=1\),\(S_{10}=33\),则数列\(\{a_n \}\)的公比是( )
A.\(8\) \(\qquad \qquad \qquad \qquad\) B.\(4\) \(\qquad \qquad \qquad \qquad\) C.\(3\) \(\qquad \qquad \qquad \qquad\) D.\(2\)
3.已知各项均为正数的等比数列\(\{a_n \}\)的前\(n\)项和是\(S_n\),且满足\(a_6\),\(3a_4\),\(-a_5\)成等差数列,则 \(\dfrac{S_4}{S_2}\)的值为( )
A.\(3\) \(\qquad \qquad \qquad \qquad\) B.\(9\) \(\qquad \qquad \qquad \qquad\) C.\(10\) \(\qquad \qquad \qquad \qquad\) D.\(13\)
4.我国古代的数学名著《九章算术》中有“衰分问题”:今有女子善织,日自倍,五日织五尺,问日织几何?其意为:一女子每天织布的尺数是前天的\(2\)倍,\(5\)天共织布\(5\)尺,问第五天织布的尺数是多少?你的答案是( )
A. \(\dfrac{5}{31}\) \(\qquad \qquad \qquad \qquad\) B.\(1\) \(\qquad \qquad \qquad \qquad\) C. \(\dfrac{5}{2}\) \(\qquad \qquad \qquad \qquad\) D. \(\dfrac{80}{31}\)
5.(多选)已知\(S_n\)是公比\(q\)的正项等比数列\(\{a_n \}\)的前\(n\)项和,若\(a_1+a_2=3\),\(a_2 a_4=16\),则下列说法正确的是( )
A.\(q=2\) \(\qquad \qquad \qquad \qquad\) B.数列\(\{S_n+1\}\)是等比数列
C.\(S_6=63\) \(\qquad \qquad \qquad \qquad\) D.数列\(\{\lg a_n\}\)是公差为\(2\)的等差数列
6.已知等比数列\(\{a_n \}\)的前\(n\)项和是\(S_n\),且\(\dfrac{S_6}{3 S_3}=\dfrac{3}{8}\),则\(\dfrac{2 a_6}{a_5+a_4}=\)\(\underline{\quad \quad}\).
7.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\(\underline{\quad \quad}\).
8.《九章算术》中有如下问题:今有蒲生一日,长三尺,莞生一日,长\(1\)尺.蒲生日自半,莞生日自倍.问几何日而长等?意思是:今有蒲第一天长高\(3\)尺,莞第一天长高\(1\)尺,以后蒲每天长高前一天的一半,莞每天长高前一天的\(2\)倍.若蒲、莞长度相等,求所需时间.
(结果精确到\(0.1\).参考数据:\(\lg 2=0.3010\),\(\lg 3=0.4771\).)
9.等比数列\(\{a_n \}\)的前\(n\)项和为\(S_n\),已知\(S_1\),\(S_3\),\(S_2\)成等差数列.
(1)求\(\{a_n \}\)的公比\(q\); \(\qquad \qquad\)(2)若\(a_1-a_3=3\),求\(S_n\).
10.设\(\{a_n \}\)是等比数列,其前\(n\)项的和为\(S_n\),且\(a_2=2\),\(S_2-3a_1=0\).
(1)求\(\{a_n \}\)的通项公式; \(\qquad \qquad\) (2)若\(S_n+a_n≥48\),求\(n\)的最小值.
11.已知数列\(\{b_n \}\)满足\(b_1=1\),\(b_{n+1} =\dfrac{1}{2} b_n\).
(1)求\(\{b_n \}\)的通项公式; \(\qquad \qquad\) (2)求\(b_2+b_4+b_6+⋯+b_{2n}\)值.
参考答案
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答案 \(A\)
解析 由\(a_1=1\),\(S_4=5S_2\),得 \(\dfrac{1-q^4}{1-q}=5 \cdot \dfrac{1-q^2}{1-q}\),即\(q^4-5q^2+4=0\),
显然\(q≠1\)所以解得\(q=-2\)(舍去)或\(q=2\),
所以\(a_5=a_1 q^4=1×2^4=16\).
故选:\(A\). -
答案 \(D\)
解析 \(\because S_n\)是等比数列\(\{a_n \}\)的前\(n\)项和,\(S_5=1\), \(S_{10}=33\),
\(\therefore \dfrac{S_{10}}{S_5}=\dfrac{1-q^{10}}{1-q^5}=1+q^5=33\),
\(\therefore q^5=32\),\(q=2\).
故选:\(D\). -
答案 \(C\)
解析 设各项均为正数的等比数列\(\{a_n \}\)的公比为\(q>0\),
\(\because\)满足\(a_6\),\(3a_4\),\(-a_5\)成等差数列,
\(\therefore 6a_4=a_6-a_5\), \(\therefore 6a_4=a_4 (q^2-q)\),
\(\therefore q^2-q-6=0\),\(q>0\).解得\(q=3\).
则\(\dfrac{S_4}{S_2}=\dfrac{\dfrac{a_1\left(3^4-1\right)}{3-1}}{\dfrac{a_1\left(3^2-1\right)}{3-1}}=3^2+1=10\).
故选:\(C\). -
答案 \(D\)
解析 由题意,该女子每天织布的尺数成等比数列\(\{a_n \}\),其公比为\(q=2\),
设第一天织布的尺数为\(a_1\),第一天织布的尺数为\(a_5\),前\(5\)天共织布为\(S_5=5\),
则\(S_5=\dfrac{a_1\left(1-2^5\right)}{1-2}=5\),解得 \(a_1=\dfrac{5}{31}\),
所以 \(a_5=a_1 q^4=\dfrac{5}{31} \times 2^4=\dfrac{80}{31}\).
故选:\(D\). -
答案 \(ABC\)
解析 因为正项等比数列\(\{a_n \}\)中,\(a_1+a_2=3\),\(a_2 a_4=16\),
所以\(\left\{\begin{array}{l} a_1+a_1 q=3 \\ a_1^2 q^4=16 \end{array}\right.\),
因为\(q>0\),解得\(q=2\)或 \(q=-\dfrac{2}{3}\)(舍),\(A\)正确;
所以\(a_1=1\), \(S_n+1=\dfrac{1-2^n}{1-2}+1=2^n\),
故数列是等比数列\(\{S_n+1\}\),\(B\)正确;
\(S_6=\dfrac{1-2^6}{1-2}=63\),\(C\)正确;
\(a_n=2^{n-1}\) ,
则\(\lg a_n=(n-1) \lg 2\),数列\(\{\lg a_n\}\)是公差为\(\lg 2\)的等差数列,\(D\)错误;
故选:\(ABC\). -
答案 \(\dfrac{1}{3}\)
解析 \(\because\)等比数列\(\{a_n \}\)中, \(\dfrac{S_6}{3 S_3}=\dfrac{3}{8}\),
显然\(q≠1\), \(\therefore \dfrac{a_1\left(1-q^6\right)}{1-q}=\dfrac{9}{8} \times \dfrac{a_1\left(1-q^3\right)}{1-q}\),
\(\therefore 1-q^6=\dfrac{9}{8}\left(1-q^3\right)\), \(\therefore 1+q^3=\dfrac{9}{8}\),\(\therefore q=\dfrac{1}{2}\),
\(\therefore \dfrac{2 a_6}{a_5+a_4}=\dfrac{2 a_1 q^5}{a_1\left(q^4+q^3\right)}=\dfrac{2 q^2}{1+q}=\dfrac{\frac{1}{2}}{\frac{3}{2}}=\dfrac{1}{3}\). -
答案 \(24\)
解析 由题意可得:\(S_8-2S_4=6\),可得:\(S_8-S_4=S_4+6\),
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 \(S_4\left(S_{12}-S_8\right)=\left(S_8-S_4\right)^2\),
综上可得: \(a_9+a_{10}+a_{11}+a_{12}=S_{12}-S_8=\dfrac{\left(S_4+6\right)^2}{S_4}=S_4+\dfrac{36}{S_4}+12 \geq 24\),
当且仅当\(S_4=6\)时等号成立.
综上可得,则\(a_9+a_{10}+a_{11}+a_{12}\)的最小值为\(24\).
故答案为:\(24\). -
答案 \(2.6\)日
解析 设蒲的长度组成等比数列\(\{a_n \}\),其\(a_1=3\),公比为\(\dfrac{1}{2}\),其前\(n\)项和为\(A_n\).
莞的长度组成等比数列\(\{b_n \}\),其\(b_1=1\),公比为\(2\),
其前\(n\)项和为\(B_n\).则 \(A_n=\dfrac{3\left(1-\dfrac{1}{n}\right)}{1-\dfrac{1}{2}}\), \(B_n=\dfrac{2^n-1}{2-1}\),
由题意可得\(\dfrac{3\left(1-\dfrac{1}{2^n}\right)}{1-\dfrac{1}{2}}=\dfrac{2^n-1}{2-1}\),化为 \(2^n+\dfrac{6}{2^n}=7\),
解得\(2^n=6\),\(2^n=1\)(舍去).
\(\therefore n=\dfrac{\lg 6}{\lg 2}=1+\dfrac{\lg 3}{\lg 2} \approx 2.6\).
\(\therefore\)估计\(2.6\)日蒲、莞长度相等. -
答案 (1) \(-\dfrac{1}{2}\);(2) \(S_n=\dfrac{8}{3}\left[1-\left(-\dfrac{1}{2}\right)^n\right]\) .
解析 (1)依题意,有\(2S_3=S_1+S_2\),
即\(a_1+(a_1+a_1 q)=2(a_1+a_1 q+a_1 q^2 )\).
由于\(a_1≠0\),故\(2q^2+q=0\).
又\(q≠0\),所以\(q=-\dfrac{1}{2}\).
(2)由已知,可得 \(a_1-a_1\left(-\dfrac{1}{2}\right)^2=3\),解得\(a_1=4\).
从而\(S_n=\dfrac{4 \times\left[1-\left(-\dfrac{1}{2}\right)^n\right]}{1-\left(-\dfrac{1}{2}\right)}=\dfrac{8}{3}\left[1-\left(-\dfrac{1}{2}\right)^n\right]\). -
答案 (1) \(a_n=2^{n-1}\) ;(2) \(6\) .
解析 (1)设等比数列的公比\(q\),
\(\because S_2-3a-1 =0\),\(\therefore a_2=2a_1=2\),\(\therefore q=2\),\(a_1=1\),
\(\therefore a_n=a_1 q^{n-1} =2^{n-1}\) .
(2) \(\because S_n=\dfrac{1-2^n}{1-2}=2^{n-1}\) ,
\(\because S_n+a_n=2^n-1+2^{n-1} =3\cdot 2^{n-1} -1≥48\),
\(\therefore 3\cdot 2^{n-1} >49\), \(\therefore 2^{n-1}>\dfrac{49}{3}\),
\(\therefore n≥6\)即\(n\)的最小值\(6\). -
答案 (1) \(b_n=\left(\dfrac{1}{2}\right)^{n-1}\);(2) \(\dfrac{2}{3}\left[1-\left(\dfrac{1}{4}\right)^n\right]\).
解析 (1)由\(b_{n+1} =\dfrac{1}{2} b_n\)得 \(\dfrac{b_n+1}{b_n}=\dfrac{1}{2}\),
所以\(\{b_n \}\)为等比数列,且首项\(b_1=1\),公比\(q=\dfrac{1}{2}\),
所以\(\{b_n \}\)的通项公式为\(b_n=\left(\dfrac{1}{2}\right)^{n-1}\).
(2)设\(a_n=b_{2n}\),则\(\dfrac{a_{n+1}}{a_n}=\dfrac{b_{2 n+2}}{b_{2 n}}=\dfrac{\left(\frac{1}{2}\right)^{2 n+1}}{\left(\frac{1}{2}\right)^{2 n-1}}=\left(\dfrac{1}{2}\right)^2=\dfrac{1}{4}\),
所以\(\{b_{2n} \}\)是首项为\(\dfrac{1}{2}\),公比\(\dfrac{1}{4}\)的等比数列,
所以 \(b_2+b_4+b_6+\cdots+b_{2 n}=\dfrac{\dfrac{1}{2}\left[1-\left(\dfrac{1}{4}\right)^n\right]}{1-\dfrac{1}{4}}=\dfrac{2}{3}\left[1-\left(\dfrac{1}{4}\right)^n\right]\).
【B组---提高题】
1.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.\(7\)
2.已知数列\(\{a_n \}\)的前\(n\)项和是\(S_n\),\(a_1=\dfrac{1}{3}\),当\(n≥2\)时,\(a_n\),\(S_n-1\),\(S_n\)成等比数列,若 \(S_m<\dfrac{19}{21}\),则\(m\)的最大值为( )
A.\(9\) \(\qquad \qquad \qquad \qquad\) B.\(11\) \(\qquad \qquad \qquad \qquad\) C.\(19\) \(\qquad \qquad \qquad \qquad\) D.\(21\)
3.(多选)设等比数列\(\{a_n \}\)的公比为\(q\),其前\(n\)项和为\(S_n\),前\(n\)项积为\(T_n\),并满足条件\(a_1>1\), \(a_{2019} a_{2020}>1\), \(\dfrac{a_{2019}-1}{a_{2020}-1}<0\),下列结论正确的是( )
A.\(S_{2019}<S_{2020}\) \(\qquad\) B.\(S_{2019} S_{2021}-1<0\) \(\qquad\)C.\(T_{2019}\)是数列\(\{T_n\}\)中的最大值 \(\qquad\) D.数列\(\{T_n\}\)无最大值
参考答案
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答案 \(A\)
解析 等比数列\(\{a_n \}\)中,公比\(q>0\);
由\(a_1 a_3=16=a_2^2\),所以\(a_2=4\),
又\(S_3=14\),
所以\(\left\{\begin{array}{l} a_1 \cdot a_3=16 \\ a_1+a_3=10 \end{array}\right.\),解得\(\left\{\begin{array}{l} a_1=2 \\ a_3=8 \end{array}\right.\)或\(\left\{\begin{array}{l} a_1=8 \\ a_3=2 \end{array}\right.\),
若\(\left\{\begin{array}{l} a_1=2 \\ a_3=8 \end{array}\right.\)时,可得\(q=2\),可得\(a_1,a_2,⋯,a_{n_0}\)的值为\(2,4,8,16…\),
不会存在\(n_0\)使得\(a_1,a_2,⋯,a_{n_0}\)的乘积最大(舍去),
若\(\left\{\begin{array}{l} a_1=8 \\ a_3=2 \end{array}\right.\)时,可得\(q=\dfrac{1}{2}\),可得\(a_1,a_2,⋯,a_{n_0}\)的值为\(8,4,2,1,\dfrac{1}{2},\dfrac{1}{4}⋯\),
观察可知存在\(n_0=4\),使得\(8×4×2×1\)的乘积最大,
综上,可得\(n_0\)的一个可能是\(4\).
故选:\(A\). -
答案 \(A\)
解析 依题意,因为当\(n≥2\)时,\(a_n\),\(S_n-1\),\(S_n\)成等比数列,
所以\((S_n-1)^2=a_n S_n\),即\((S_n-1)^2=(S_n-S_{n-1} )S_n\),
即\(\dfrac{1}{S_n-1}-\dfrac{1}{S_{n-1}-1}=-1\),
所以\(\left\{\dfrac{1}{S_n-1}\right\}\)成等差数列,所以 \(\dfrac{1}{S_n-1}=-n-\dfrac{1}{2}\),即\(S_n=\dfrac{2 n-1}{2 n+1}\),
若\(S_m<\dfrac{19}{21}\),即\(\dfrac{2 m-1}{2 m+1}<\dfrac{19}{21}\),解得\(m<10\),
所以\(m\)的最大值为\(9\).
故选:\(A\). -
答案 \(AC\)
解析 等比数列\(\{a_n \}\)的公比为q,其前\(n\)项和为\(S_n\),前\(n\)项积为\(T_n\),
并满足条件\(a_1>1\), \(a_{2019} a_{2020}>1\), \(\dfrac{a_{2019}-1}{a_{2020}-1}<0\),
\(\therefore a_{2019}>1\),\(0<a_{2020}<1\), \(\therefore 0<q<1\).
根据\(a_1>1\),\(0<q<1\),可知等比数列\(\{a_n \}\)为正项的递减数列.
即\(a_1>a_2>⋯>a_{2019}>1>a_{2020}>⋯>0\).
\(\because S_{2020}-S_{2019}=a_{2020}>0\),
\(\therefore S_{2019}<S_{2020}\),故选项\(A\)正确;
\(\because S_{2019}=a_1+a_2+⋯+a_{2019}>1\),
\(\therefore S_{2019} S_{2021}=S_{2019}\left(S_{2019}+a_{2020}+a_{2021}\right)\)\(=S_{2019}^2+S_{2019}\left(a_{2020}+a_{2021}\right)>S_{2019}^2>1\).
即\(S_{2019} S_{2021}-1>0\).故选项\(B\)错误;
根据\(a_1>a_2>⋯>a_{2019}>1>a_{2020}>⋯>0\).可知
\(T_{2019}\)是数列\(\{T_n\}\)中的最大项,故选项\(C\)正确、选项\(D\)错误.
故选:\(AC\).
【C组---拓展题】
1.设\(m\)个正数\(a_1\),\(a_2\),\(…\),\(a_m\)\((m≥4,m∈N^*)\)依次围成一个圆圈.其中\(a_1\),\(a_2\),\(a_3\),\(…\),\(a_{k-1}\) ,\(a_k\)\((k<m,k∈N^*)\)是公差为\(d\)的等差数列,而\(a_1\),\(a_m\),\(a_{m-1}\) ,\(…\),\(a_{k+1}\),\(a_k\)是公比为\(2\)的等比数列.
(1)若\(a_1=d=2\),\(k=8\),求数列\(a_1\),\(a_2\),\(…\),\(a_m\)的所有项的和\(S_m\);
(2)若\(a_1=d=2\),\(m<2015\),求\(m\)的最大值;
(3)是否存在正整数\(k\),满足\(a_1+a_2+a_3+⋯+a_{k-1} +a_k=3\)\((a_{k+1}+a_{k+2}+⋯+a_{m-1} +a_m)\)?
若存在,求出\(k\)值;若不存在,请说明理由.
参考答案
- 答案 (1) \(84\);(2) \(1033\);(3) \(k=4\).
解析 (1)依题意\(a_k=16\),
故数列\(a_1\),\(a_2\),\(…\),\(a_m\)即为\(2,4,6,8,10,12,14,16,8,4\)共\(10\)个数,
此时\(m=10\),\(S_m=84\).
(2)由数列\(\{a_n \}\)满足\(a_1=d=2\),是首项为\(2\)、公差为\(2\)的等差数列知,\(a_k=2k\),
而\(a_1\),\(a_m\),\(a_{m-1}\) ,\(…\),\(a_{k+1}\),\(a_k\)是首项为\(2\)、公比为\(2\)的等比数列知, \(a_k=2^{m+2-k}\),
故有 \(2 k=2^{m+2-k}\), \(k=2^{m+1-k}\),即\(k\)必是\(2\)的整数次幂,
由 \(k \cdot 2^k=2^{m+1}\)知,要使\(m\)最大,\(k\)必须最大,
又\(k<m<2015\),故\(k\)的最大值\(2^{10}\),
从而 \(2^{10} \cdot 2^{1024}=2^{m+1}\),\(m\)的最大值是\(1033\).
(3)由数列\(\{a_n \}\)是公差为\(d\)的等差数列知,\(a_k=a_1+(k-1)d\),
而\(a_1\),\(a_m\),\(a_{m-1}\) ,\(…\),\(a_{k+1}\),\(a_k\)是公比为\(2\)的等比数列 \(a_k=a_1 \cdot 2^{m+1-k}\),
故 \(a_1+(k-1) d=a_1 \cdot 2^{m+1-k}\), \((k-1) d=a_1\left(2^{m+1-k}-1\right)\),
又\(a_1+a_2+a_3+\cdots+a_{k-1}+a_k=3\left(a_{k+1}+a_{k+2}+\cdots+a_{m-1}+a_m\right)\),\(a_m=2a_1\),
则\(k a_1+\dfrac{1}{2} k(k-1) d=3 \times 2 a_1 \times \dfrac{1-2^{m-k}}{1-2}\),
即\(k a_1+\dfrac{1}{2} k\left[a_1\left(2^{m+1-k}-1\right)\right]=3 \times 2 a_1\left(2^{m-k}-1\right)\),
则\(\dfrac{1}{2} k \cdot 2^{m+1-k}+\dfrac{1}{2} k=6\left(2^{m-k}-1\right)\),
即\(k \cdot 2^{m+1-k}+k=6 \times 2^{m+1-k}-12\),
显然\(k≠6\),则 \(2^{m+1-k}=\dfrac{k+12}{6-k}=-1+\dfrac{18}{6-k}\)
所以\(k<6\),将\(k=1,2,3,4,5\)一一代入验证知,
当\(k=4\)时,上式右端为\(8\),等式成立,此时\(m=6\),
综上可得:当且仅当\(m=6\)时,存在\(k=4\)满足等式.