4.3.2 等比数列的综合应用
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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,\(n∈ N^* )\)
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,\(n≥2)⇒\{a_n\}\)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.
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 \(S_n= \begin{cases}n a_1 & (q=1) \\ \dfrac{a_1\left(1-q^n\right)}{1-q} & (q \neq 1)\end{cases}\)
(由错位相减法可证)
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\(\{λa_n\}\)\((λ\)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\(\{a_n b_n\}\)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\(a_k\) , \(a_{k+m}\) , \(a_{k+2m}\),\(…\)\((k ,m∈N^*)\)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,\(S_n=0\))
基本方法
【题型1】 等比数列的基本运算
【典题1】 若公比为\(2\)的等比数列\(\{a_n \}\)的前\(n\)项和为\(S_n\),且\(a_2\),\(9\),\(a_5\)成等差数列,则\(S_{20}=\) ( )
A. \(2^{22}-1\) \(\qquad \qquad \qquad\) B. \(2^{21}-1\) \(\qquad \qquad \qquad\) C. \(2^{20}-1\) \(\qquad \qquad \qquad\) D. \(2^{19}-1\)
解析 设等比数列\(\{a_n \}\)的首项为\(a_1\),已知公比\(q=2\),且\(a_2\),\(9\),\(a_5\)成等差数列,
\(18=a_2+a_5=2a_1+16a_1\),解得\(a_1=1\).
\(\therefore S_{20}=\dfrac{1 \times\left(1-2^{20}\right)}{1-2}=2^{20}-1\).
故选:\(C\).
【典题2】 已知数列\(\{a_n \}\)是公差不为零的等差数列,\(\{b_n \}\)为等比数列,且\(a_1=b_1=1\),\(a_2=b_2\),\(a_4=b_3\),
设\(c_n=a_n+b_n\),则数列\(\{c_n \}\)的前\(10\)项和为\(\underline{\quad \quad}\) .
解析 设等差数列\(\{a_n \}\)的公差为\(d≠0\),等比数列\(\{b_n\}\)的公比为\(q\),
\(\because a_1=b_1=1\),\(a_2=b_2\),\(a_4=b_3\),
\(\therefore 1+d=q\),\(1+3d=q^2\),\(d≠0\),解得\(d=1\),\(q=2\).
\(\therefore a_n=1+n-1=n\),\(b_n=2^{n-1}\).
\(\therefore c_n=a_n+b_n=n+2^{n-1}\).
则数列\(\{c_n \}\)的前\(10\)项和\(\dfrac{10 \times(1+10)}{2}+\dfrac{2^{10}-1}{2-1}=1078\).
【巩固练习】
1.设正项等比数列\(\{a_n \}\)的前\(n\)项和为\(S_n\),若\(2S_3=3a_2+8a_1\),则公比\(q=\)( )
A.\(2\)\(\qquad \qquad \qquad \qquad\) B.\(-\dfrac{3}{2}\) \(\qquad \qquad \qquad \qquad\) C.\(2\)或\(-\dfrac{3}{2}\) \(\qquad \qquad \qquad \qquad\) D.\(2\)或 \(-\dfrac{3}{2}\)
2.已知等比数列\(\{a_n \}\)的前\(n\)项和为\(S_n\),且\(2S_3\),\(3S_5\),\(4S_6\)成等差数列,则数列\(\{a_n \}\)的公比q=$( )
A.\(1\)或\(\dfrac{1}{2}\) \(\qquad \qquad\) B.\(-1\)或\(-\dfrac{1}{2}\) \(\qquad \qquad\) C.\(-1\)或\(2\) \(\qquad \qquad\) D.\(1\)或\(-2\)
3.设正项等比数列\(\{a_n \}\)的前\(n\)项和为\(S_n\),若\(2S_3=3a_2+8a_1\),\(S_8=2S_7+2\),则\(a_2=\)( )
A.\(4\) \(\qquad \qquad \qquad \qquad\) B.\(3\) \(\qquad \qquad \qquad \qquad\) C.\(2\) \(\qquad \qquad \qquad \qquad\) D.\(1\)
参考答案
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答案 \(A\)
解析 由\(2S_3=3a_2+8a_1\),有\(2(a_1+a_2+a_3 )=3a_2+8a_1\),即\(2a_3-a_2-6a_1=0\),
由等比数列的通项公式得\(2a_1 q^2-a_1 q-6a_1=0\),
即\(2q^2-q-6=0\),解得\(q=2\)或\(q=-\dfrac{3}{2}\),
由数列为正项等比数列, \(\therefore q=2\).
故选:\(A\). -
答案 \(B\)
解析 \(\because2S_3\),\(3S_5\),\(4S_6\)成等差数列,\(\therefore 6S_5=2S_3+4S_6\),
即\(6(a_1+a_2+a_3+a_4+a_5 )=2(a_1+a_2+a_3 )+4(a_1+a_2+a_3+a_4+a_5+a_6 )\),
整理得\(6a_4+6a_5=4a_4+4a_5+4a_6\),即\(4a_6-2a_5-2a_4=0\),
\(\because a_4≠0\),\(\therefore 2q^2-q-1=0\),解得\(q=1\)或\(-\dfrac{1}{2}\),
故选:\(B\). -
答案 \(A\)
解析 设正项等比数列\(\{a_n \}\)的公比为\(q\),
\(\because 2S_3=3a_2+8a_1\),
\(\therefore 2(a_1+a_2+a_3 )=3a_2+8a_1\),即\(6a_1+a_2-2a_3=0\),
\(\therefore 6a_1+a_1 q-2a_1 q^2=0\),
\(\because a_1>0\),
\(\therefore 6+q-2q^2=0\),解得\(q=2\)或\(q=-\dfrac{3}{2}\)(舍去),\(\therefore q=2\),
\(\because S_8=2S_7+2\),\(\therefore S_7+a_8=2S_7+2\),\(\therefore a_8=S_7+2\),
\(\therefore a_1 q^7=\dfrac{a_1\left(1-q^7\right)}{1-q}+2\),
\(\because q=2\), 解得\(a_1=2\),
\(\therefore 128a_1=127a_1+2\),解得\(a_1=2\),
\(\therefore a_2=a_1 q=4\).
故选:\(A\).
【题型2】 等比数列的基本性质的运用
【典题1】 (多选)在递增的等比数列\(\{a_n \}\)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.\(q=1\) \(\qquad \qquad\) B.数列\(\{S_n+2\}\)是等比数列\(\qquad \qquad\) C.\(S_8=510\) \(\qquad \qquad\) D.数列\(\{\lg a_n \}\)是公差为\(2\)的等差数列
解析 由题意,根据等比中项的性质,可得\(a_2 a_3=a_1 a_4=32>0\),\(a_2+a_3=12>0\),
故\(a_2>0\),\(a_3>0\).
根据根与系数的关系,可知\(a_2\),\(a_3\)是一元二次方程\(x^2-12x+32=0\)的两个根.
解得\(a_2=4\),\(a_3=8\),或\(a_2=8\),\(a_3=4\).
\(\because\)等比数列\(\{a_n \}\)是递增数列,\(\therefore q>1\).
\(\therefore a_2=4\),\(a_3=8\)满足题意.
\(\therefore q=2\), \(a_1=\dfrac{a_2}{q}=2\).故选项\(A\)不正确.
\(a_n=a_1⋅q^{n-1}=2^n\).
\(\because S_n=\dfrac{2\left(1-2^n\right)}{1-2}=2^{n+1}-2\), \(\therefore S_n+2=2^{n+1}=4 \cdot 2^{n-1}\).
\(\therefore\)数列\(\{S_n+2\}\)是以\(4\)为首项,\(2\)为公比的等比数列.故选项\(B\)正确.
\(S_8=2^{8+1}-2=512-2=510\).故选项\(C\)正确.
\(\because \lg a_n=\lg 2^n=n\lg 2\).
\(\therefore\) 数列\(\{\lg a_n \}\)是公差为\(\lg 2\)的等差数列.故选项\(D\)不正确.
故选:\(BC\).
【典题2】 等比数列\(\{a_n \}\)的前\(n\)项和为\(S_n\),前\(n\)项积为\(T_n\),\(a_4-a_2=6\),\(a_5-a_3=12\),当 \(\dfrac{T_n}{\left(S_n+1\right)^{\frac{7}{2}}}\)最小时,\(n\)的值为( )
A.\(3\) \(\qquad \qquad \qquad \qquad\) B.\(4\) \(\qquad \qquad \qquad \qquad\) C.\(5\) \(\qquad \qquad \qquad \qquad\) D.\(6\)
解析 等比数列\(\{a_n \}\)的前\(n\)项和为\(S_n\),前\(n\)项积为\(T_n\),\(a_4-a_2=6\),\(a_5-a_3=12\),
当\(q≠1\)时,\(a_2 (q^2-1)=6\),\(a_2 q^3-a_2 q=12\),
两式相除得\(q=2\),
将\(q=2\)代入\(a_2 (q^2-1)=6\),解得\(a_2=2\),
\(\therefore a_1=1\), \(\therefore S_n=\dfrac{1-2^n}{1-2}=2^n-1\),
\(T_n=1×2×2^2×2^3×⋯×2^{n-1}=2^{\frac{n(n-1)}{2}}\),
\(\therefore \dfrac{T_n}{\left(S_n+1\right)^{\frac{7}{2}}}=\dfrac{2^{\frac{n(n-1)}{2}}}{2^{\frac{7 n}{2}}}=2^{\frac{n^2-8 n}{2}}\),
\(\therefore n=4\)时, \(\dfrac{n^2-8 n}{2}\)取得最小值,即当 \(\dfrac{T_n}{\left(S_n+1\right)^{\frac{7}{2}}}\)最小时,\(n\)的值为\(4\).
故选:\(B\).
【巩固练习】
1.已知等比数列\(\{a_n \}\)的前\(n\)项和为\(S_n\),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.\(\left[0,1\right]\)
2.公比为\(q\)的等比数列\(\{a_n \}\),其前\(n\)项和为\(S_n\),前\(n\)项积为\(T_n\),满足\(a_1>1\),\(a_{2021}\cdot a_{2022}>1\), \(\dfrac{a_{2021}-1}{a_{2022}-1}<0\).则下列结论正确的是( )
A.\(q<0\) \(\qquad \qquad\) B.\(a_{2021}\cdot a_{2023}>1\) \(\qquad \qquad\)C.\(S_n\)的最大值为\(S_{2023}\) \(\qquad \qquad\) D.\(T_n\)的最大值为\(T_{2021}\)
3.(多选)已知数列\(\{a_n \}\)是公比为\(q\)的等比数列,其前\(n\)项和为\(S_n\),则下列结论正确的是( )
A.若数列\(\{a_n \}\)是正项等比数列,则数列\(\{\lg a_n \}\)是等差数列
B.若\(a_3=\dfrac{3}{2}\), \(S_3=\dfrac{9}{2}\),则\(q=-\dfrac{1}{2}\)
C.若\(a_3=1\),\(a_7=16\),则\(a_5=4\)
D.若 \(S_6-S_4=6(S_4-S_2 )\),则\(q^2=6\)
4.已知等比数列\(\{a_n \}\)的前\(n\)项和为\(S_n\),公比\(q>0\),\(a_1=1\), \(a_{12}=9 a_{10}\),若数列\(\{t+S_n \}\)为等比数列,则实数\(t=\)\(\underline{\quad \quad}\).
5.已知等比数列\(\{a_n \}\),\(a_n>0\),\(a_1=256\),\(S_3=448\),\(T_n\)为数列\(\{a_n \}\)的前\(n\)项乘积,则当\(T_n\)取得最大值时,\(n=\)\(\underline{\quad \quad}\).
参考答案
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答案 \(B\)
解析 设等比数列\(\{a_n \}\)的公比为\(q\),\(\because a_1+2a_2=0\), \(S_3=\dfrac{3}{4}\),
\(\therefore a_1 (1+2q)=0\),\(a_1 (1+q+q^2 )=\dfrac{3}{4}\),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\(\dfrac{1}{2}\),
\(\therefore\left\{\begin{array}{l} a \leqslant \dfrac{1}{2} \\ a+2 \geqslant 1 \end{array}\right.\),\(\therefore -1⩽a⩽\dfrac{1}{2}\),
故选:\(B\). -
答案 \(D\)
解析 等比数列\(\{a_n \}\)中,\(a_1>1\),\(a_{2021}\cdot a_{2022}>1\), \(\dfrac{a_{2021}-1}{a_{2022}-1}<0\).
所以\(a_{2021}>1\),\(0<a_{2022}<1\),\(0<q<1\),\(A\)错误;
故当\(n=2021\)时,\(T_n\)取得最大值,\(D\)正确;
由于数列各项都为正数,和没有最大值,\(C\)错误;
由题意得,\(0<a_{2022}<1\),\(0<a_{2023}<1\),
所以\(a_{2021}\cdot a_{2023}=a_{2022}^2<1\),\(B\)错误.
故选:\(D\). -
答案 \(AC\)
解析 对于\(A\),不妨设正项等比数列\(\{a_n \}\)的通项公式\(a_n=a_1 q^{n-1}\),
因为\(\lg a_n={n-1}\lg q+\lg a_1\)是\(n\)的一次式,\(\{\lg a_n \}\)是等差数列,
所以\(A\)正确,
对于\(B\),因为\(S_3=a_1+a_2+a_3=a_3 (q^{-2}+q^{-1}+1)\),
所以 \(q^{-2}+q^{-1}+1=3\),即\(2q^2-q-1=0\),解得\(q=1\)或\(q=-\dfrac{1}{2}\),
所以\(B\)不正确,
对于\(C\),若\(a_3=1\),\(a_7=16\),
则\(a_5^2=a_3 a_7=16\),注意到 \(\dfrac{a_5}{a_3}=q^2>0\),
所以\(a_5=4\),所以\(C\)正确,
对于\(D\),由\(S_6-S_4=6(S_4-S_2 )\)得\(a_5+a_6=6(a_3+a_4 )\),
所以\(q^2 (a_3+a_4 )=6(a_3+a_4 )\),
当\(a_3+a_4=0\)时,\(q=-1\),\(q^2=1\),所以\(D\)不正确.
故选:\(AC\). -
答案 \(\dfrac{1}{2}\)
解析 因为 \(a_{12}=9 a_{10}\),所以\(a_1\cdot q^{11}=9a_1\cdot q^9\),即\(q^2=9\),
因为\(q>0\),所以\(q=3\),
所以 \(S_n=\dfrac{a_1\left(1-q^n\right)}{1-q}=\dfrac{3^n-1}{2}\),
所以\(S_1=1\),\(S_2=4\),\(S_3=13\),
因为数列\(\{t+S_n \}\)为等比数列,
所以\((t+S_2 )^2=(t+S_1 )\cdot (t+S_3 )\),
所以\((t+4)^2=(t+1)\cdot (t+13)\),解得\(t=\dfrac{1}{2}\). -
答案 \(8\)或\(9\)
解析 设等比数列\(\{a_n \}\)的公比为\(q\),\(\because a_n>0\),\(\therefore q>0\).
\(\because a_1=256\),\(S_3=448\),\(\therefore 256(1+q+q^2 )=448\),解得\(q=\dfrac{1}{2}\).
\(\therefore a_n=256 \times\left(\dfrac{1}{2}\right)^{n-1}=2^{9-n}\).
\(T_n=2^8 \cdot 2^7 \cdot \ldots \ldots \cdot 2^{9-n}=2^{8+7+\cdots+9-n}=2^{\frac{n(8+9-n)}{2}}=2^{-\left[\left(n-\frac{17}{2}\right)^2 -\frac{289}{4}\right]}\).
\(\therefore\)当\(n=8\)或\(9\)时,\(T_n\)取得最大值时.
【题型3】等比数列解答题
【典题1】 设数列\(\{a_n \}\)的前\(n\)项和为\(S_n\),满足\(2S_n=a_{n+1}-2^{n+1}+1\),且\(a_1\),\(a_2+5\),\(a_3\)成等差数列.
(1)求\(a_1\),\(a_2\)的值;
(2)求证:\(\{a_n+2^n \}\)是等比数列.并求数列\(\{a_n \}\)的通项公式;
(3)求数列\(\{a_n+2^n+2n\}\)的前\(n\)项和.
解析 (1)\(\because a_1\),\(a_2+5\),\(a_3\)成等差数列, \(\therefore a_1+a_3=2(a_2+5)\)①,
\(\because\) 数列\(\{a_n \}\)的前\(n\)项和为\(S_n\),满足\(2S_n=a_{n+1}-2^{n+1}+1\),
当\(n=1\)时,\(2a_1=a_2-3\),②
当\(n=2\)时,\(2(a_1+a_2 )=a_3-7\),③
\(\therefore\)联立①②③解得,\(a_1=1\),\(a_2=5\),\(a_3=19\).
(2)证明:由\(2S_n=a_{n+1}-2^{n+1}+1\),①得\(2S_{n-1}=a_n-2^n+1\),②,
两式相减得\(2a_n=a_{n+1}-a_n-2^n (n⩾2)\),
\(\dfrac{a_{n+1}+2^{n+1}}{a_n+2^n}=\dfrac{3 a_n+2^n+2^{n+1}}{a_n+2^n}=3(n \geqslant 2)\).
\(\because \dfrac{a_2+2^2}{a_1+2}=3\),\(\therefore \{a_n+2^n \}\)是首项为\(3\),公比为\(3\)的等比数列.
\(\therefore a_{n+1}+2^{n+1}=3(a_n+2^n )\),
又\(a_1=1\),\(a_1+2^1=3\),
\(\therefore a_n+2^n=3^n\),即\(a_n=3^n-2^n\).
(3) 由(2)可知\(a_n=3^n-2^n\),设\(b_n=a_n+2^n+2n=3^n+2n\),前\(n\)项和为\(T_n\),
则\(T_n=b_1+b_2+⋯+b_n=(3+3^2+⋯+3^n )+(2+4+⋯+2n)\)
\(=\dfrac{3\left(1-3^n\right)}{1-3}+\dfrac{n(2+2 n)}{2}=\dfrac{3^{n+1}}{2}+n^2+n-\dfrac{3}{2} .\).
【典题2】 已知数列\(\{a_n \}\)是各项为正数的等比数列,且\(a_2=4\),\(a_3 a_4 a_5=2^{12}\).数列\(\{b_n\}\)是单调递增的等差数列,且\(b_2 b_3=15\),\(b_1+b_4=8\),
(1)求数列\(\{a_n \}\)与数列\(\{b_n\}\)的通项公式;
(2)求数列\(\{a_n b_n\}\)的前\(n\)项和\(T_n\).
解析 (1)设正项等比数列\(\{a_n \}\)的公比为\(q(q>0)\),
由 \(a_3 a_4 a_5=2^{12}\),得\(a_4^3=2^{12}\),得\(a_4=2^4\),
又\(a_2=4\), \(\therefore q^2=\dfrac{a_4}{a_2}=\dfrac{2^4}{4}=4\),则\(q=2\).
\(\therefore a_n=a_2 q^{n-2}=4 \cdot 2^{n-2}=2^n\);
在等差数列\(\{b_n\}\)中,由等差数列的性质可得,\(b_2+b_3=b_1+b_4=8\),
又\(b_2 b_3=15\),且数列\(\{b_n\}\)是单调递增数列,
解得\(b_2=3\),\(b_3=5\),
则公差\(d=b_3-b_2=5-3=2\).
\(\therefore b_n=b_2+(n-2)×2=3+2n-4=2n-1\).
(2)\(\because a_n b_n=(2n-1) 2^n\).
\(\therefore T_n=1\cdot 2^1+3\cdot 2^2+5\cdot 2^3+⋯+(2n-1)\cdot 2^n\),
\(2T_n=1\cdot 2^2+3\cdot 2^3+5\cdot 2^4+⋯+(2n-3)\cdot 2^n+(2n-1)\cdot 2^{n+1}\).
\(\therefore -T_n=2+2(2^2+2^3+⋯+2^n)-(2n-1)\cdot 2^{n+1}\)
\(=2+2 \cdot \dfrac{4\left(1-2^{n-1}\right)}{1-2}-(2 n-1) \cdot 2^{n+1}=-(2 n-3) 2^{n+1}-6\).
\(\therefore T_n=(2n-3)\cdot 2^{n+1}+6\).
【巩固练习】
1.已知等比数列\(\{a_n \}\)满足\(a_2 a_3=2a_4=32\).
(1)求\(\{a_n \}\)的通项公式;
(2)记\(\{a_n \}\)的前\(n\)项和为\(S_n\),证明:\(n⩾2\)时,\(a_n^2>S_n+5\).
2.已知数列\(\{a_n \}\)满足\(a_1=3\),\(a_2=5\),且\(2a_{n+2}=3a_{n+1}-a_n\),\(n∈N^*\).
(1)设\(b_n=a_{n+1}-a_n\),求证:数列\(\{b_n\}\)是等比数列;
(2)若数列\(\{a_n \}\)满足\(a_n⩽m(n∈N^* )\),求实数\(m\)的取值范围.
3.设数列\(\{a_n \}\)的前\(n\)项和为\(S_n\),\(a_1=1\),\(S_{n+1}=4a_n+2(n∈N^* )\)
(1)若\(b_n=a_{n+1}-2a_n\),求\(b_n\);
(2)若 \(c_n=\dfrac{1}{a_{n+1}-2 a_n}\),求\(\{c_n \}\)的前\(6\)项和\(T_6\);
(3)若 \(d_n=\dfrac{a_n}{2^n}\) ,求数列\(\{d_n \}\)的通项.
参考答案
-
答案 (1) \(a_n=2^n\) ;(2) 略 .
解析 (1)解:设等比数列\(\{a_n \}\)的公比为\(q\),
因为\(a_2 a_3=2a_4=32\),所以\(a_1^2 q^3=2a_1 q^3=32\),
又等比数列\(\{a_n \}\)中\(a_1\)和\(q\)均不为\(0\),所以\(a_1=q=2\),
故数列\(\{a_n \}\)的通项公式为\(a_n=a_1\cdot q^{n-1}=2\cdot 2^{n-1}=2^n\).
(2)证明:由(1)可得\(S_n=\dfrac{a_1\left(1-q^n\right)}{1-q}=\dfrac{2\left(1-2^n\right)}{1-2}=2^{n+1}-2\),
因为\(n⩾2\)时,\(2^n>3\),
所以\(a_n^2-S_n-5=(2^n )^2-2^{n+1}-3=(2^n )^2-2\cdot 2^n-3=(2^n+1)(2^n-3)>0\),
所以\(n⩾2\)时,\(a_n^2>S_n+5\). -
答案 (1) 略 ;(2)\(\left[7, +∞\right)\).
解析 (1)证明:由题知,\(2a_{n+2}=3a_{n+1}-a_n\),
即\(b_{n+1}=\dfrac{1}{2} b_n\),且\(b_1=a_2-a_1=5-3=2\),
则数列\(\{b_n\}\)是以\(2\)为首项,\(\dfrac{1}{2}\)为公比的等比数列.
(2)解:由(1)知 \(b_n=a_{n+1}-a_n=\dfrac{1}{2^{n-2}}\) ,
则当\(n⩾2\)时,
其前\(n-1\)项和\(S_{n-1}=a_2-a_1+a_3-a_2+⋯+a_n-a_{n-1}\)
\(=a_n-a_1=\dfrac{2-\dfrac{1}{2^{n-2}}}{1-\dfrac{1}{2}}=4-\dfrac{1}{2^{n-3}}\),
则\(a_n=7-\left(\dfrac{1}{2}\right)^{n-3}\) ,\(n⩾2\),且\(a_1=3\)也满足通项,
则由指数函数单调性知,\(a_n=7-\dfrac{1}{2^{n-3}}<7\),
若满足\(a_n⩽m(n∈N^* )\),则\(m⩾7\),
即实数\(m\)的取值范围是\([7, +∞)\). -
答案 (1) \(b_n=3\cdot 2^{n-1}\);(2) \(\dfrac{61}{96}\);(3) \(d_n=\dfrac{3 n}{4}-\dfrac{1}{4}\) .
解析 (1)\(\because a_1=1\),\(S_{n+1}=4a_n+2(n∈N^* )\),
\(\therefore S_{n+2}=4a_{n+1}+2\),
\(\therefore a_{n+2}=S_{n+2}-S_{n+1}=4(a_{n+1}-a_n )\),
\(\therefore a_{n+2}-2a_{n+1}=2(a_{n+1}-a_n )\),即\(b_{n+1}=2b_n\),
\(\therefore \{b_n \}\)是公比为\(2\)的等比数列,且\(b_1=a_2-2a_1\)
\(\because a_1=1\),\(a_2+a_1=S_2\),即\(a_2+a_1=4a_1+2\),
\(\therefore a_2=3a_1+2=5\),\(\therefore b_1=5-2=3\),
\(\therefore b_n=3\cdot 2^{n-1}\).
(2) \(c_n=\dfrac{1}{a_{n+1}-2 a_n}=\dfrac{1}{b_n}=\dfrac{1}{3 \cdot 2^{n-1}}\), \(\therefore c_1=\dfrac{1}{3}\),
\(\therefore c_n=\dfrac{1}{3} \cdot\left(\dfrac{1}{2}\right)^{n-1}\),
\(\therefore \{c_n \}\)是首项为\(\dfrac{1}{3}\),公比为\(\dfrac{1}{2}\)的等比数列,
\(\therefore T_6=\dfrac{\dfrac{1}{3}\left[1-\left(\dfrac{1}{2}\right)^6\right]}{1-\dfrac{1}{2}}=\dfrac{2}{3}\left(1-\dfrac{1}{64}\right)=\dfrac{61}{96}\).
(3) \(\because d_n=\dfrac{a_n}{2^n}\),\(b_n=3\cdot 2^{n-1}\),
\(\therefore d_{n+1}-d_n=\dfrac{a_{n+1}}{2^{n+1}}-\dfrac{a_n}{2^n}=\dfrac{a_{n+1}-2 a_n}{2^{n+1}}=\dfrac{b_n}{2^{n+1}}=\dfrac{3 \times 3^{n-1}}{2^{n+1}}=\dfrac{3}{4}\),
\(\therefore \{d_n \}\)是等差数列, \(d_n=\dfrac{3 n}{4}-\dfrac{1}{4}\).
分层练习
【A组---基础题】
1.记单调递增的等比数列\(\{a_n \}\)的前\(n\)项和为\(S_n\),若\(a_2+a_4=10\),\(a_2 a_3 a_4=64\),则( )
A.\(S_{n+1}-S_n=2^{n+1}\) \(\qquad \qquad\) B.\(a_n=2^n\) \(\qquad \qquad\) C.\(S_n=2^n-1\) \(\qquad \qquad\) D.\(S_n=2^{n-1}-1\)
2.设等比数列\(\{a_n \}\)的前\(n\)项和为\(S_n\),公比为\(q\),且\(S_3\),\(S_9\),\(S_6\)成等差数列,则\(8q^3\)等于( )
A.\(-2\) \(\qquad \qquad \qquad \qquad\) B.\(-4\) \(\qquad \qquad \qquad \qquad\) C.\(2\) \(\qquad \qquad \qquad \qquad\) D.\(4\)
3.设\(S_n\)为等比数列\(\{a_n \}\)的前\(n\)项和,若\(a_n>0\),\(a_1=\dfrac{1}{2}\),\(S_n<2\),则\(\{a_n \}\)的公比的取值范围是( )
A.\(\left(0,\dfrac{3}{4}\right]\) \(\qquad \qquad \qquad\) B.\(\left(0,\dfrac{2}{3}\right]\) \(\qquad \qquad \qquad\) C.\(\left(0,\dfrac{3}{4}\right)\) \(\qquad \qquad \qquad\) D.\(\left(0,\dfrac{2}{3}\right)\)
4.在等比数列\(\{a_n \}\)中,\(a_1=-9\),\(a_5=-1\),记 \(T_n=a_1 a_3 a_5 \ldots a_{2 n-1}(n=1,2, \cdots)\),则数列\(\{T_n \}\) ( )
A.有最大项,有最小项 \(\qquad \qquad \qquad \qquad\) B.有最大项,无最小项
C.无最大项,有最小项 \(\qquad \qquad \qquad \qquad\) D.无最大项,无最小项
5.(多选)在公比为\(q\)的等比数列\(\{a_n \}\)中,\(S_n\)是数列\(\{a_n \}\)的前\(n\)项和,若\(a_1=1\),\(a_5=27a_2\),则下列说法正确的是( )
A.\(q=3\) \(\qquad \qquad \qquad \qquad \qquad \qquad\) B.数列\(\{S_n+2\}\)是等比数列
C.\(S_5=121\) \(\qquad \qquad \qquad \qquad \qquad \quad\) D.\(2\lg a_n=\lg a_{n-2}+\lg a_{n+2} (n⩾3)\)
6.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 \(\dfrac{S_8-2 S_4}{S_6-S_4-S_2}=\)\(\underline{\quad \quad}\).
7.如图,在\(Rt△ABC\)内有一系列的正方形,它们的边长依次为\(a_1\),\(a_2\),\(…\),\(a_n\),\(…\),若\(AB=a\),\(BC=2a\),
则所有正方形的面积的和为\(\underline{\quad \quad}\).
8.已知等比数列\(\{a_n \}\)的公比\(q>0\),且\(a_1=1\),\(4a_3=a_2 a_4\).
(1)求公比\(q\)和\(a_3\)的值;
(2)若\(\{a_n \}\)的前\(n\)项和为\(S_n\),求证 \(\dfrac{S_n}{a_n}<2\).
9.在数列\(\{a_n \}\)中,\(a_1=4\)且对于任意的自然数\(n∈N^+\)都有\(a_{n+1}=2(a_n-n+1)\).
(1)证明数列\(\{a_n-2n\}\)是等比数列.
(2)求数列\(\{a_n \}\)的通项公式及前\(n\)项和为\(S_n\).
10.设数列\(\{a_n \}\)的前\(n\)项和为\(S_n\),\(a_1=3\)且\(a_{n+1}=2S_n+3\),数列\(\{b_n\}\)为等差数列,且公差\(d>0\),\(b_1+b_2+b_3=15\).
(1)求数列\(\{a_n \}\)的通项公式;
(2)若\(\dfrac{a_1}{3}+b_1\), \(\dfrac{a_2}{3}+b_2\), \(\dfrac{a_3}{3}+b_3\)成等比数列,求数列\(\{b_n\}\)的前\(n\)项和\(T_n\);
(3)对第(2)小题的\(T_n\),当\(T_n+16⩾λn\)对任意的\(n∈N^*\)恒成立,求\(λ\)的最大值.
参考答案
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答案 \(C\)
解析 \(\because a_2 a_3 a_4=64\),\(\therefore a_3^3=64\),解得\(a_3=4\).
又\(a_2+a_4=10\), \(\therefore \dfrac{4}{q}+4 q=10\),
化为\(2q^2-5q+2=0\),解得\(q=2\)或\(\dfrac{1}{2}\).
\(q=2\)时,\(a_1=1\);\(q=\dfrac{1}{2}\)时,\(a_1=16\).
又等比数列\(\{a_n \}\)是单调递增,取\(q=2\),\(a_1=1\).
\(\therefore a_n=2^{n-1}\).
\(\therefore S_n=\dfrac{2^n-1}{2-1}=2^n-1\),\(S_{n+1}-S_n=2^{n+1}-1-(2^n-1)=2^n\).
因此只有\(C\)正确.
故选:\(C\). -
答案 \(B\)
解析 \(\because S_3\),\(S_9\),\(S_6\)成等差数列,\(\therefore 2S_9=S_3+S_6\),
\(\therefore (S_9-S_6 )+(S_9-S_3 )=0\),
即\((a_7+a_8+a_9 )+(a_7+a_8+a_9 )+(a_4+a_5+a_6 )=0\),
\(\therefore 2q^3 (a_4+a_5+a_6 )+(a_4+a_5+a_6 )=0\),
\(\because a_4+a_5+a_6=a_4 (1+q+q^2 )≠0\),
\(\therefore q^3=-\dfrac{1}{2}\), \(\therefore 8q^3=-4\).
故选:\(B\). -
答案 \(A\)
解析 设等比数列\(\{a_n \}\)的公比为\(q\),则\(q≠1\).
\(\because a_n>0\),\(a_1=\dfrac{1}{2}\),\(S_n<2\),
\(\therefore \{a_n \}\)是递减数列,\(\dfrac{1}{2}×q^{n-1}>0\), \(\dfrac{\dfrac{1}{2}\left(1-q^n\right)}{1-q}<2\),
\(\therefore 1>q>0\)且\(1⩽4-4q\),解得\(0<q⩽\dfrac{3}{4}\).
综上,\(\{a_n \}\)的公比的取值范围是\(\left(0,\dfrac{3}{4}\right]\).
故选:\(A\). -
答案 \(A\)
解析 设等比数列\(\{a_n \}\)的公比为\(q\),
则 \(\dfrac{a_1}{a_1}=q^4=\dfrac{1}{9}\),所以\(q^2=\dfrac{1}{3}\),
设数列\(\{b_n\}\)为等比数列\(\{a_n \}\)的奇数项\(a_1\),\(a_3\),\(a_5\),…,\(a_{2n-1} (n=1,2,…)\),
则数列\(\{b_n\}\)是以\(-9\)为首项,\(\dfrac{1}{3}\)为公比的等比数列,
则\(b_n=-9 \times\left(\dfrac{1}{3}\right)^{n-1}=-\left(\dfrac{1}{3}\right)^{n-3}\),
所以 \(T_n=a_1 a_3 a_5 \cdots a_{2 n-1}=b_1 b_2 b_3 \cdots b_n\),
当\(n⩾4\)时,\(|b_n |<1\),当\(1⩽n⩽3\)时,\(|b_n |⩾1\),
当\(n\)为奇数时,\(T_n<0\),因为\(b_3=-1\),所以\(T_n⩾T_3=-27\),
当\(n\)为偶数时,\(T_n>0\),因为\(b_3=-1\),所以\(T_n⩽T_2=27\),
综上所述,数列\(\{T_n \}\)有最大项\(T_2=27\)和最小项\(T_3=-27\).
故选:\(A\). -
答案 \(ACD\)
解析 根据题意,依次分析选项:
对于\(A\),等比数列\(\{a_n \}\)中,\(S_n\)是数列\(\{a_n \}\)的前\(n\)项和,
若\(a_1=1\),\(a_5=27a_2\),则 \(q^3=\dfrac{a_5}{a_2}=27\),解可得\(q=3\),\(A\)正确;
对于\(B\),由于\(q=3\),\(a_1=1\),则 \(S_n=\dfrac{a_1\left(1-q^n\right)}{1-q}=\dfrac{3^n-1}{2}\),
则 \(S_n+2=\dfrac{3^n+1}{2}\),数列\(\{S_n+2\}\)不是等比数列,\(B\)错误;
对于\(C\),由于 \(S_n=\dfrac{a_1\left(1-q^n\right)}{1-q}=\dfrac{3^n-1}{2}\),则\(S_5=121\),\(C\)正确;
对于\(D\),等比数列\(\{a_n \}\)各项都大于\(0\),
又由等比数列的性质可知,\(a_{n-2}\cdot a_{n+2}=a_n^2\),
所以\(2\lg a_n=\lg a_{n+2}+\lg a_{n-2}\),\(D\)正确.
故选:\(ACD\). -
答案 \(10\)
解析 设等比数列\(\{a_n \}\)的公比为\(q\),
因为\(S_4=a_1+a_2+a_3+a_4=(1+q)(a_1+a_3 )\),所以\(q=3\),
则\(\dfrac{S_8-2 S_4}{S_6-S_4-S_2}=\dfrac{\left(S_8-S_4\right)-S_4}{\left(S_6-S_2\right)-S_4}\)\(=\dfrac{q^4 S_4-S_4}{q^2 S_4-S_4}=\dfrac{q^4-1}{q^2-1}=q^2+1=10\). -
答案 \(\dfrac{4}{5} a^2\)
解析 根据题意可知 \(\dfrac{A B}{B C}=\dfrac{1}{2}\),可得 \(a_1=\dfrac{2}{3} a\),
依次计算\(a_2=\dfrac{2}{3} a_1\),\(a_3=\dfrac{2}{3} a_2\),\(…\)是公比为\(\dfrac{2}{3}\)的等比数列,
正方形的面积:依次 \(S_1=\dfrac{4}{9} a^2\), \(S_2=\dfrac{4}{9} a_1^2\),\(…\)边长依次为\(a_1\),\(a_2\),…,\(a_n\),
正方形的面积构成是公比为 \(\dfrac{4}{9}\)的等比数列.
所有正方形的面积的和\(S_n=\dfrac{S_1}{1-q}=\dfrac{\dfrac{4}{9} a^2}{1-\dfrac{4}{9}}=\dfrac{4}{5} a^2\).
故答案为:\(\dfrac{4}{5} a^2\) -
答案 (1) \(q=2\),\(a_3=4\);(2) 略 .
解析 (1)解:\(\because\)等比数列\(\{a_n \}\)的公比\(q>0\),且\(a_1=1\),\(4a_3=a_2 a_4\).
\(\therefore 4q^2=q^4\),解得\(q=2\),\(\therefore a_3=4\).
(2)证明:\(a_n=2^{n-1}\), \(S_n=\dfrac{2^n-1}{2-1}=2^n-1\),
\(\therefore \dfrac{s_n}{a_n}-2=\dfrac{2^n-1}{2^{n-1}}-2=2-\dfrac{1}{2^{n-1}}-2<0\),
\(\therefore \dfrac{S_n}{a_n}<2\). -
答案 (1)略;(2)\(S_n=2^{n+1}-2+n^2+n\) .
解析 (1)\(\because a_{n+1}=2(a_n-n+1)\)
\(\therefore \dfrac{a_{n+1}-2(n+1)}{a_n-2 n}=\dfrac{2\left(a_n-n+1\right)-2(n+1)}{a_n-2 n}=\dfrac{2\left(a_n-2 n\right)}{a_n-2 n}=2\),
\(\therefore\)数列\(\{a_n-2n\}\)是以\(a_1-2=2\)为首项,以\(2\)为公比的等比数列
(2)由(1)可得\(a_n-2n=2\cdot 2^{n-1}=2^n\)
\(\therefore a_n=2^n+2n\),
\(\therefore S_n=\dfrac{2-2^{n+1}}{1-2}+\dfrac{(2+2 n) n}{2}=2^{n+1}-2+n^2+n\). -
答案 (1) \(a_n=3^n\) ;(2) \(T_n=n^2+2n\) ;(3) \(10\).
解析 (1)由\(a_{n+1}=2S_n+3\),得\(a_n=2S_{n-1}+3(n⩾2)\)
相减得:\(a_{n+1}-a_n=2(S_n-S_{n-1} )\),即\(a_{n+1}=3a_n\),
\(\because\)当\(n=1\)时,\(a_2=2a_1+3=9\), \(\therefore \dfrac{a_2}{a_1}=3\),
\(\therefore\)数列\(\{a_n \}\)是等比数列,
\(\therefore a_n=3\cdot 3^{n-1}=3^n\).
(2)\(\because b_1+b_2+b_3=15\),\(b_1+b_3=2b_2\), \(\therefore b_2=5\)
由题意,\(\dfrac{a_1}{3}+b_1\), \(\dfrac{a_2}{3}+b_2\), \(\dfrac{a_3}{3}+b_3\)成等比数列, \(\therefore\left(\dfrac{a_2}{3}+b_2\right)^2=\left(\dfrac{a_1}{3}+b_1\right)\left(\dfrac{a_3}{3}+b_3\right)\),
设\(b_1=5-d\),\(b_3=5+d\),\(\therefore 64=(5-d+1)(5+d+9)\),
\(\therefore d^2+8d-20=0\),得\(d=2\)或\(d=-10\)(舍去)
故\(T_n=3 n+\dfrac{n(n-1)}{2} \cdot 2=n^2+2 n\).
(3)由题意, \(\lambda \leqslant n+\dfrac{16}{n}+2\),
\(\because n+\dfrac{16}{n} \geqslant 2 \sqrt{n \cdot \dfrac{16}{n}}=8\),
\(\therefore λ\)的最大值为\(8+2=10\).
【B组---提高题】
1.首项为\(729\)的等比数列\(\{a_n \}\)满足 \(a_n=3^{b_n}\),记数列\(\{b_n\}\)的前\(n\)项和为\(S_n\),若\(∀n∈N^*\),当\(n≠4\)时,\(S_n<S_4\),则数列\(\{a_n \}\)的公比\(q\)的取值范围为( )
A. \(\left(\dfrac{1}{9}, \dfrac{\sqrt{3}}{9}\right)\) \(\qquad \qquad\)B. \(\left(\dfrac{1}{3}, \dfrac{\sqrt{3}}{3}\right)\) \(\qquad \qquad\) C. \(\left(\dfrac{1}{9}, \dfrac{\sqrt{3}}{3}\right)\) \(\qquad \qquad\) D. \(\left(\dfrac{1}{9}, \dfrac{1}{3}\right)\)
2.(多选)已知数列\(\{a_n \}\)满足\(a_1=1\), \(a_{n+1}=\lg \left(10^{a_n}+9\right)+1\),其前\(n\)项和为\(S_n\),则下列结论中正确的有( )
A.\(\{a_n \}\)是递增数列 \(\qquad \qquad \qquad \qquad \qquad\) B.\(\{a_n+10\}\)是等比数列
C.\(2a_{n+1}>a_n+a_{n+2}\) \(\qquad \qquad \qquad \qquad\)D.\(S_n<\dfrac{n(n+3)}{2}\)
3.已知等比数列\(\{a_n \}\)的首项为\(\dfrac{3}{2}\),公比为\(-\dfrac{1}{2}\),前\(n\)项和为\(S_n\),且对任意的\(n∈N^*\),都有 \(A \leqslant 2 S_n-\dfrac{1}{S_n} \leqslant B\)恒成立,则\(B-A\)的最小值为\(\underline{\quad \quad}\).
4.已知数列\(\{a_n \}\)是等差数列,其前\(n\)项和为\(A_n\),\(a_7=15\),\(A_7=63\);数列\(\{b_n\}\)的前\(n\)项和为\(B_n\),\(2B_n=3b_n-3(n∈N^* )\).
(1)求数列\(\{a_n \}\),\(\{b_n\}\)的通项公式;
(2)求数列\(\left\{\dfrac{1}{A_n}\right\}\)的前\(n\)项和为\(S_n\);
(3)求证: \(\sum_{k=1}^n \dfrac{a_k}{B_k}<2\).
参考答案
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答案 \(A\)
解析 等比数列\(\{a_n \}\)的公比为\(q\),因为\(a_n=3^{b_n}\),所以\(q>0\),
则当\(n⩾2\)时, \(\dfrac{a_n}{a_{n-1}}=\dfrac{3^{b_n}}{3^{b_{n-1}}}=3^{b_n-b_{n-1}}=q\),
所以 \(b_n-b_{n-1}=\log _3 q\),又 \(729=3^6=3^{b_1}\),即\(b_1=6\),
所以数列\(\{b_n\}\)是以\(6\)为首项,以\(\log_3q\)为公差的等差数列,
等差数列\(\{b_n\}\)前\(n\)项和\(S_n\)中,\(S_4\)是唯一的最大值,
则\(\left\{\begin{array}{l} b_4>0 \\ b_5<0 \end{array}\right.\),即\(\begin{cases}6+3 \log _3 & q>0 \\ 6+4 \log _3 & q<0\end{cases}\),解得\(\dfrac{1}{9}<q<\dfrac{\sqrt{3}}{9}\),
故选:\(A\). -
答案 \(ACD\)
解析 因为 \(a_{n+1}=\lg \left(10^{a_n}+9\right)+1\),所以 \(10^{a_{n+1}}=10\left(10^{a_n}+9\right)\),
所以 \(10^{a_{n+1}}+10=10\left(10^{a_n}+10\right)\),
令 \(b_n=10^{a_n}+10\),则\(b_{n+1}=10b_n\),
即\(\{b_n\}\)是以\(10\)为公比的等比数列,\(b_1=20\),
故\(b_n=2×10^n\),
所以\(a_n=\lg (2×10^n-10)\)是递增数列,但不是等比数列,\(A\)正确,\(B\)错误;
因为\(2a_{n+1}=\lg (4×10^{2n+2}+100-40×10^{n+1} )\),
\(a_n+a_{n+2}=\lg (2×10^n-10)(20×10^n-10)\)
\(=\lg \left[4×10^{2n+2}+100-20(10^n+10^{n+2} )\right]\),
又 \(10^n+10^{n+2}>2 \sqrt{10^{2 n+2}}=2 \times 10^{n+1}\),
所以\(a_n+a_{n+2}<2a_{n+1}\),\(C\)正确;
令\(c_n=n+1\),则其前\(n\)项和为 \(\dfrac{n(n+3)}{2}\),
而\(a_n=\lg (2×10^n-10)<\lg (2×10^n )<\lg (10×10^n )=n+1=c_n\),
故 \(S_n<\dfrac{n(n+3)}{2}\),\(D\)正确.
故选:\(ACD\). -
答案 \(\dfrac{13}{6}\)
解析 \(S_n=\dfrac{\dfrac{3}{2}\left[1-\left(-\dfrac{1}{2}\right)^n\right]}{1-\left(-\dfrac{1}{2}\right)}=1-\left(-\dfrac{1}{2}\right)^n\).
\(2 S_n-\dfrac{1}{S_n}=2\left[1-\left(-\dfrac{1}{2}\right)^n\right]-\dfrac{1}{1-\left(-\dfrac{1}{2}\right)^n}\).
\(n=2k(k∈N^* )\)时, \(2 S_n-\dfrac{1}{S_n}=2\left[1-\left(\dfrac{1}{2}\right)^n\right]-\dfrac{1}{1-\left(\dfrac{1}{2}\right)^n}\),关于\(n\)单调递增,
\(n=2\)时,取得最小值 \(2 S_2-\dfrac{1}{S_2}=\dfrac{1}{6}\); \(n\rightarrow+\infty\)时, \(2 S_n-\dfrac{1}{S_n} \rightarrow 1\).
\(n=2k-1(k∈N^* )\)时, \(2 S_n-\dfrac{1}{S_n}=2\left[1+\left(\dfrac{1}{2}\right)^n\right]-\dfrac{1}{1+\left(\dfrac{1}{2}\right)^n}\),关于\(n\)单调递减,\(n=1\)时,取得最大值 \(2 S_1-\dfrac{1}{S_1}=\dfrac{7}{3}\);
\(n\rightarrow+\infty\)时, \(2 S_n-\dfrac{1}{S_n} \rightarrow 1\).
对任意的\(n∈N^*\),都有 \(A \leqslant 2 S_n-\dfrac{1}{S_n} \leqslant B\)恒成立,
则\(B-A\)的最小值为 \(\dfrac{7}{3}-\dfrac{1}{6}=\dfrac{13}{6}\).
故答案 为: \(\dfrac{13}{6}\). -
答案 (1)\(a_n=2n+1\),\(b_n=3^n\) ;(2) \(S_n=\dfrac{3}{4}-\dfrac{2 n+3}{2(n+1)(n+2)}\) ;(3)略 .
解析 (1)\(\because\)数列\(\{a_n \}\)是等差数列,其前n项和为\(A_n\),\(a_7=15\),\(A_7=63\),
\(\therefore\left\{\begin{array}{l} a_7=a_1+6 d=15 \\ A_7=7 a_1+\dfrac{7 \times 6}{2} d=63 \end{array} \right.\),解得\(a_1=3\),\(d=2\),
\(\therefore\) 数列\(\{a_n \}\)的通项公式为\(a_n=3+{n-1}×2=2n+1\),
\(\because\)数列\(\{b_n\}\)的前n项和为\(B_n\),\(2B_n=3b_n-3(n∈N^* )\),
\(\therefore n⩾2\)时,\(2b_n=2(B_n-B_{n-1} )=3b_n-3-(3b_{n-1}-3)\),
化为\(b_n=3b_{n-1}\),
\(n=1\),\(2b_1=3b_1-3\),解得\(b_1=3\),
\(\therefore \{b_n \}\)是以\(3\)为首项,\(3\)为公比的等比数列,
\(\therefore\)数列\(\{b_n\}\)的通项公式为\(b_n=3^n\);
(2)\(\because\)数列\(\{a_n \}\)的通项公式为 \(A_n=3 n+\dfrac{n(n-1)}{2} \cdot 2=n^2+2 n\),
\(\therefore \dfrac{1}{A_n}=\dfrac{1}{n^2+2 n}=\dfrac{1}{n(n+2)}=\dfrac{1}{2}\left(\dfrac{1}{n}-\dfrac{1}{n+2}\right)\),
\(\therefore\)数列 \(\left\{\dfrac{1}{A_n}\right\}\)的前 项和 \(S_n=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{3}-\dfrac{1}{5}+\cdots+\dfrac{1}{n-1}-\dfrac{1}{n+1}+\dfrac{1}{n}-\dfrac{1}{n+2}\right)\),
即 \(S_n=\dfrac{1}{2}\left(1+\dfrac{1}{2}-\dfrac{1}{n+1}-\dfrac{1}{n+2}\right)=\dfrac{3}{4}-\dfrac{2 n+3}{2(n+1)(n+2)}\);
(3)证明:\(\because \{b_n \}\)是以\(3\)为首项,\(3\)为公比的等比数列,
\(\therefore\) 数列\(\{b_n\}\)的前\(n\)项和为 \(B_n=\dfrac{3\left(1-3^n\right)}{1-3}=\dfrac{3}{2}\left(3^n-1\right)\),
\(\therefore \dfrac{a_k}{B_k}=\dfrac{2}{3} \cdot \dfrac{2 k+1}{3^k-1} \leqslant \dfrac{2}{3} \times \dfrac{2 k+1}{2 \cdot 3^{k-1}}=\dfrac{2 k+1}{3^k}\),
令\(T_k\)为 \(\left\{\dfrac{2 k+1}{3^k}\right\}\)的前\(k\)项和,则\(T_k=\dfrac{3}{3^1}+\dfrac{5}{3^2}+\cdots+\dfrac{2 k-1}{3^{k-1}}+\dfrac{2 k+1}{3^k}\),
两式相减得\(\dfrac{2}{3} T_k=1+2\left(\dfrac{1}{3^2}+\dfrac{1}{3^3}+\cdots+\dfrac{1}{3^k}\right)-\dfrac{2 k+1}{3^{k+1}}\)
\(=1+2 \times \dfrac{\dfrac{1}{9}\left(1-\dfrac{1}{3^{k-1}}\right)}{1-\dfrac{1}{3}}-\dfrac{2 k+1}{3^{k+1}}=\dfrac{4}{3}-\dfrac{2 k+4}{3^{k+1}}\),
所以 \(T_k=2-\dfrac{k+2}{3^k}\),
故 \(\sum_{k=1}^n \dfrac{a_k}{B_k} \leqslant T_k=2-\dfrac{n+2}{3^n}<2\).
【C组---拓展题】
1.抛物线\(y^2=4px(p>0)\)的准线与\(x\)轴的交点为\(M\),过点\(M\)作直线交抛物线于\(A\),\(B\)两点.
(1)求线段\(AB\)中点的轨迹方程;
(2)若\(p=\dfrac{1}{2}\),线段\(AB\)的垂直平分线交对称轴于点\(N(x_N,0)\),求证:\(x_N>\dfrac{3}{2}\);
(3)若\(p=\dfrac{1}{2}\),直线\(l\)的斜率依次取\(\dfrac{1}{2}\),\(\left(\dfrac{1}{2}\right)^2\),\(…\), \(\left(\dfrac{1}{2}\right)^n\)时,线段\(AB\)的垂直平分线与抛物线对称轴的交点依次是\(N_1\),\(N_2\),…,\(N_n\),求\(S=\dfrac{1}{\left|N_1 N_2\right|}+\dfrac{1}{\left|N_2 N_3\right|}+\cdots+\dfrac{1}{\left|N_n N_{n+1}\right|}\).
参考答案
- 答案 (1) \(y^2=2px+2p^2,(p>0,x>0)\);(2)略;(3)\(\dfrac{1}{9}\left(1-\dfrac{1}{4^n}\right)\).
解析 (1)抛物线的准线方程为\(x=-p\),\(\therefore M(-p,0)\),
设\(l\)方程为\(y=k(x+p)(k≠0)\),\(A(x_1,y_1 )\),\(B(x_2,y_2 )\),\(AB\)中点\(P(x_0,y_0)\),
由\(\left\{\begin{array}{c} y=k(x+p) \\ y^2=4 p x \end{array}\right.\)得\(k^2 x^2+(2pk^2-4p)x+k^2 p^2=0 \quad (*)\),
则\(x_1+x_2=\dfrac{4 p-2 p k^2}{k^2}\), \(\therefore x_0=\dfrac{x_1+x_2}{2}=\dfrac{2 p}{k^2}-p\) ①,
又\(y_0=k\left(x_0+p\right)=\dfrac{2 p}{k}\) ②,
由①,②消\(k\)后得\(y_0^2=2px_0+2p^2,(p>0,x_0>0)\),
即\(AB\)中点\(P\)轨迹方程为\(y^2=2px+2p^2,(p>0,x>0)\).
(2)证明:设\(AB\)中点\(P(x_0,y_0)\),由(1)可知 \(x_0=\dfrac{2 p}{k^2}-p=\dfrac{1}{k^2}-\dfrac{1}{2}\),
又 \(y_0=k\left(x_0+p\right)=k\left(\dfrac{1}{k^2}-\dfrac{1}{2}+\dfrac{1}{2}\right)=\dfrac{1}{k}\),
\(\therefore P\left(\dfrac{1}{k^2}-\dfrac{1}{2}, \dfrac{1}{k}\right)\)
则线段\(AB\)的垂直平分线方程为\(y-\dfrac{1}{k}=-\dfrac{1}{k}\left[x-\left(\dfrac{1}{k^2}-\dfrac{1}{2}\right)\right]\),
令\(y=0\), \(x_N=\dfrac{1}{k^2}+\dfrac{1}{2}\),
由(1)中的方程(*),可知\(\Delta=\left(2 p k^2-4 p\right)^2-4 k^4 p^2=16 p^2-16 p k^2>0\),
\(\therefore 0<k^2<1\), \(\therefore \dfrac{1}{k^2}+\dfrac{1}{2}>\dfrac{3}{2}\),则 \(x_N>\dfrac{3}{2}\).
(3)设\(N_n (x_n,0)\),
当直线\(l\)的斜率 \(k_n=\left(\dfrac{1}{2}\right)^n\)时,由(2)可知 \(x_n=\dfrac{1}{k_n^2}+\dfrac{1}{2}=4^n+\dfrac{1}{2}\),
\(\because\left|N_n N_{n+1}\right|=\left|x_{n+1}-x_n\right|=\left|4^{n+1}+\dfrac{1}{2}-\left(4^n+\dfrac{1}{2}\right)\right|=3 \cdot 4^n\),
\(\therefore \dfrac{1}{\left|N_n N_{n+1}\right|}=\dfrac{1}{3} \cdot\left(\dfrac{1}{4}\right)^n\),
\(\therefore \dfrac{1}{\left|N_n N_{n+1}\right|}\)是以\(\dfrac{1}{12}\)为首项,以\(\dfrac{1}{4}\)为公比的等比数列,
\(S=\dfrac{1}{\left|N_1 N_2\right|}+\dfrac{1}{\left|N_2 N_3\right|}+\cdots+\dfrac{1}{\left|N_n N_{n+1}\right|}=\dfrac{\dfrac{1}{12}\left(1-\left(\dfrac{1}{4}\right)^n\right)}{1-\dfrac{1}{4}}=\dfrac{1}{9}\left(1-\dfrac{1}{4^n}\right)\).