4.2.1 等差数列的概念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
定义
如果一个数列从第二项起,每一项与它的前一项的差等于同一个常数,那么这个数列叫做等差数列,
这个常数叫做等差数列的公差,记为\(d\) .
代数形式:\(a_n-a_{n-1}=d\) (\(n≥2,d\)是常数)
解释
(1)公差是每一项减前一项,常数指的是与\(n\)无关;
(2)公差\(d\in R\),当\(d=0\)时,数列为常数列;当\(d>0\)时,数列为递增数列;当\(d<0\)时,数列为递减数列
(3)\(a_n-a_{n-1}=2(n≥2)⟹\{a_n\}\)是公差为\(2\)的等差数列;
\(\qquad\)\(a_{n+1}-a_n=-3⟹\{a_n\}\)是公差为\(-3\)的等差数列;
\(\qquad\)\(a_{n+1}-a_n=3n⟹\{a_n\}\)不是等差数列.
【例】以下数列是等差数列的 .
(1)\(2,,4,8,16\);\(\qquad \qquad\)(2)\(1,3,5,7,9\);\(\qquad \qquad\)(3)数列\(\{a_n \}\)满足\(a_n-a_{n-1}=1(n≥2)\).
答案 (2)(3).
等差中项
若\(a\),\(b\),\(c\)成等差数列,则\(b\)称\(a\)与\(c\)的等差中项,则 \(b=\dfrac{a+c}{2}\).
证明
若\(a\),\(b\),\(c\)成等差数列,由等差数列的定义可得\(b-a=c-b\),则 \(b=\dfrac{a+c}{2}\).
【例】 若\(x+4\)是\(5\)和\(x^2\)的等差中项,则\(x=\)\(\underline{\quad \quad}\) .
解 依题意得\(2(x+4)=5+x^2\),解得\(x=-1\)或\(3\).
通项公式
等差数列\(\{a_n \}\)的首项为\(a_1\),公差为\(d\),则\(a_n=a_1+(n-1) d\). (由定义与累加法可得)
解释
(1)证明 若等差数列\(\{a_n \}\)的首项为\(a_1\),公差为\(d\),
由等差数列的定义可得,\(a_{n+1}-a_n=d\),
所以\(a_2-a_1=d\),\(a_3-a_2=d\),\(a_4-a_3=d\),\(…\),\(a_n-a_{n-1}=d(n≥2)\),
把以上\(n-1\)项等式累加可得\(a_n-a_1=(n-1) d(n≥2)\),
当\(n=1\)时,上式为\(a_1=a_1+(1-1)d=a_1\),即上式当\(n=1\)时也成立,
故\(a_n=a_1+(n-1)d(n\in N^*)\).
等差数列的通项公式由等差数列的定义证明,以上证明方法为累加法.
(2)从函数的角度看等差数列的通项公式.
由等差数列的通项公式\(a_n=a_1+(n-1)d\)可得\(a_n=d\cdot n+(a_1-d)\),
当\(d≠0\)时,\(a_n\)是关于\(n\)的一次函数;当\(d=0\)时,\(a_n=a_1\)是常数列.
(3)由两点确定一条直线的性质可以得出,已知等差数列的任意两项可以确定这个等差数列.若已知等差数列的通项公式,可以写出数列中的任意一项.
(4)等差数列\(\{a_n \}\)的通项公式\(a_n=a_1+(n-1) d\)中共含有四个变数,即\(a_1\),\(d\),\(n\),\(a_n\),如果知道了其中的任意三个数,就可以由通项公式求出第四个数,这一求未知量的过程我们通常称之为“知三求一”.
【例】已知等差数列\(\{a_n \}\)中,首项\(a_1=4\),公差\(d=-2\),则通项公式\(a_n\)等于( )
A.\(4-2n\) \(\qquad \qquad \qquad \qquad\) B.\(2n-4\) \(\qquad \qquad \qquad \qquad\) C.\(6-2n\) \(\qquad \qquad \qquad \qquad\) D.\(2n-6\)
答案 \(C\)
证明一个数列是等差数列的方法
① 定义法: \(a_{n+1}-a_n=d\)\((d\)是常数,\(n\in N^*)\)\(⟹a_n\)是等差数列;
② 中项法: \(2a_{n+1}=a_n+a_{n+2} (n\in N^*)⟹a_n\)是等差数列.
基本方法
【题型1】 等差数列的判定与证明
【典题1】 已知数列\(\{a_n \}\)的通项公式为\(a_n=4-2n\),求证:数列\(\{a_n \}\)是等差数列.
证明 \(\because a_n=4-2n\),\(\therefore a_{n+1}=4-2{n+1}=2-2n\).
\(\therefore a_{n+1}-a_n=(2-2n)-(4-2n)=-2\).
\(\therefore \{a_n \}\)是等差数列.
点拨
- 证明等差数列的方法:定义法\(a_{n+1}-a_n=d(d\)是常数,\(n\in N^*)\);
2.若数列的通项公式\(a_n=kn+b(k,b\)是常数\()\),则该数列是等差数列.
【典题2】已知数列\(\{a_n \}\)中,\(a_1=4\),\(a_n=a_{n-1}+2^{n-1}+3(n≥2)\),证明数列\(\{a_n-2^n\}\)是等差数列.
证明 \((a_n-2^n )-(a_{n-1}-2^{n-1} )= a_n-a_{n-1}-2^{n-1}=3(n≥2,n\in N^*)\).
所以\(\{a_n-2^n\}\)是等差数列.
【巩固练习】
1.若数列\(\{a_n \}\)的通项公式为\(a_n=2n+1\),则此数列是( )
A.公差为\(2\)的等差数列 \(\qquad \qquad \qquad \qquad\) B.公差为\(5\)的等差数列
C.首项为\(5\)的等差数列 \(\qquad \qquad \qquad \qquad\) D.公差为\(n\)的等差数列
2.若数列\(\{a_n \}\)的通项公式为\(a_n=10+\lg2^n\),求证数列\(\{a_n \}\)为等差数列.
3.数列\(\{a_n \}\)满足 \(a_n=6-\dfrac{9}{a_{n-1}}\left(n \in \boldsymbol{N}^*, n \geq 2\right)\),求证:数列 \(\left\{\dfrac{1}{a_n-3}\right\}\)是等差数列.
参考答案
-
答案 \(A\)
解析 由\(a_n=2n+1\),得,公差\(a_1=3\)\(d=a_n-a_{n-1}=(2n+1)-[2{n-1}-1]=2\).
\(\therefore\) 此数列是公差为\(2\),首项为\(3\)的等差数列.
故选:\(A\). -
证明 因为\(a_n=10+\lg2^n=10+n\lg2\),
所以\(a_{n+1}=10+(n+1)\lg2\).
所以\(a_{n+1}-a_n=[10+(n+1)\lg2]-(10+n\lg2)=\lg2(n\in N^* )\).
所以数列\(\{a_n \}\)为等差数列. -
证明 数列\(\{a_n \}\)满足 \(a_n=6-\dfrac{9}{a_{n-1}}\left(n \in \boldsymbol{N}^*, n \geq 2\right)\).
\(\therefore \dfrac{1}{a_{n-3}}-\dfrac{1}{a_{n-1}-3}=\dfrac{a_{n-1}}{3 a_{n-1}-9}-\dfrac{1}{a_{n-1}-3}=\dfrac{a_{n-1}-3}{3\left(a_{n-1}-3\right)}=\dfrac{1}{3}\),
\(\therefore\)数列 \(\left\{\dfrac{1}{a_n-3}\right\}\)是等差数列.
【题型2】 等差数列的通项公式
【典题1】 已知数列\(\{a_n \}\)中,\(a_3=2\),\(a_7=1\).若 \(\left\{\dfrac{1}{a_n}\right\}\)为等差数列,则\(a_5=\)\(\underline{\quad \quad}\).
解析 设等差数列 \(\left\{\dfrac{1}{a_n}\right\}\)的公差为\(d\),
则 \(\dfrac{1}{a_7}=\dfrac{1}{a_3}+4 d\),即 \(1=\dfrac{1}{2}+4 d\),解得 \(d=\dfrac{1}{8}\).
则 \(\dfrac{1}{a_5}=\dfrac{1}{a_3}+2 d=\dfrac{1}{2}+\dfrac{1}{4}=\dfrac{3}{4}\),解得 \(a_5=\dfrac{4}{3}\).
【典题2】已知等差数列\(\{a_n \}\)中,公差\(d>0\),且\(a_2\),\(a_5\)是\(x^2-12x+27=0\)的两根,则\((a_3+a_4 )^2-a_7=\)\(\underline{\quad \quad}\) .
解析 \(\because a_2\),\(a_5\)是\(x^2-12x+27=0\)的两根,
\(\therefore\left\{\begin{array}{l}
a_2=3 \\
a_5=9
\end{array}\right.\)或 \(\left\{\begin{array}{l}
a_2=9 \\
a_5=3
\end{array}\right.\),
又\(\because\)公差\(d>0\), \(\therefore\left\{\begin{array}{l}
a_2=3 \\
a_5=9
\end{array}\right.\),
则 \(\left\{\begin{array}{c}
a_1+d=3 \\
a_1+4 d=9
\end{array}\right.\),解得 \(\left\{\begin{array}{l}
a_1=1 \\
\mathrm{~d}=2
\end{array}\right.\),
\(\therefore a_3=5\),\(a_4=7\),\(a_7=13\),
\(\therefore (a_3+a_4 )^2-a_7=12^2-13=131\).
【巩固练习】
1.在数列\(\{a_n \}\)中,\(a_1=2\),\(a_{n+1}=a_n+4\),若\(a_n=2022\),则\(n=\)( )
A.\(508\) \(\qquad \qquad \qquad \qquad\) B.\(507\) \(\qquad \qquad \qquad \qquad\) C.\(506\) \(\qquad \qquad \qquad \qquad\) D.\(505\)
2.等差数列\(1,-1,-3,…,-89\)的项数是( )
A.\(92\) \(\qquad \qquad \qquad \qquad\) B.\(47\) \(\qquad \qquad \qquad \qquad\) C.\(46\) \(\qquad \qquad \qquad \qquad\) D.\(45\)
3.已知\(\{a_n \}\)是公差为\(1\)的等差数列,且\(a_2^2=a_1 a_5\),则\(a_1=\)( )
A.\(1\) \(\qquad \qquad \qquad \qquad\) B.\(\dfrac{1}{2}\) \(\qquad \qquad \qquad \qquad\) C.\(\dfrac{1}{4}\) \(\qquad \qquad \qquad \qquad\) D.\(2\)
4.已知等差数列\(\{a_n \}\)中,\(a_2=-3\),\(a_3=-5\),则\(a_9=\)( )
A.\(-10\) \(\qquad \qquad \qquad \qquad\) B.\(-17\) \(\qquad \qquad \qquad \qquad\) C.\(-19\) \(\qquad \qquad \qquad \qquad\) D.\(-21\)
5.已知正项数列\(\{a_n \}\)的首项为\(1\),\(\{a_n^2\}\)是公差为\(3\)的等差数列,则使得\(a_n>6\)成立的\(n\)的最小值为\(\underline{\quad \quad}\) .
6.若一个三角形三边长成公差为\(2\)的等差数列,且最大角为\(120°\),则这个三角形的面积为\(\underline{\quad \quad}\) .
参考答案
-
答案 \(C\)
解析 因为数列\(\{a_n \}\)中,\(a_1=2\),\(a_{n+1}=a_n+4\),
即\(a_1=2\),\(a_{n+1}-a_n=4\),
所以数列\(\{a_n \}\)是以\(2\)为首项,以\(4\)为公差的等差数列,
\(a_n=2+4(n-1)=2022\),则\(n=506\).
故选:\(C\). -
答案 \(C\)
解析 \(a_1=1\),\(d=(-1)-1=-2\),
故\(a_n=a_1+(n-1)d=3-2n\),
令\(-89=3-2n\),解得\(n=46\). -
答案 \(B\)
解析 由\(\{a_n \}\)是公差为\(1\)的等差数列,且\(a_2^2=a_1 a_5\),
得\((a_1+1)^2=a_1 (a_1+4)\),
所以\(2a_1-1=0\),解得\(a_1=\dfrac{1}{2}\).
故选:\(B\). -
答案 \(B\)
解析 依题意得 \(\left\{\begin{array}{c} a_1+d=-3 \\ a_1+2 d=-5 \end{array}\right.\),解得\(a_1=-1\),\(d=-2\)
\(\therefore a_9=a_1+8d=-1+(-2)×8=-17\),
故选:\(B\). -
答案 \(13\)
解析 \(\because\)正项数列\(\{a_n \}\)的首项为\(1\),\(\{a_n^2\}\)是公差为\(3\)的等差数列,
\(\therefore\) 依题意得,\(a_n^2=1+3(n-1)=3n-2\),
故 \(a_n=\sqrt{3 n-2}\).令 \(\sqrt{3 n-2}>6\),得\(3n-2>36\),解得\(n>\dfrac{38}{3}\),
\(\because n\in N^*\),\(\therefore\) 使得\(a_n>6\)成立的\(n\)的最小值为\(13\). -
答案 \(\dfrac{15 \sqrt{3}}{4}\)
解析 由题意可设,三条边长分别为\(a\),\(a+2\),\(a+4\),
则 \(\left\{\begin{array}{l} a+a+2>a+4 \\ a>0 \end{array}\right.\),解得\(a>2\),
三角形最大角为\(120°\),则其最长的边的长度为\(a+4\),
\(\cos 120^{\circ}=\dfrac{a^2+(a+2)^2-(a+4)^2}{2 a \cdot(a+2)}=-\dfrac{1}{2}\),
化简整理可得\(a^2-a-6=0\),解得\(a=3\)或\(a=-2\)(舍去),
则该三角形的三条边长分别为\(3\),\(5\),\(7\),
故这个三角形的面积为\(\dfrac{1}{2} \times 3 \times 5 \times \sin 120^{\circ}=\dfrac{15 \sqrt{3}}{4}\).
【题型3】实际应用问题
【典题1】 梯子的最高一级宽\(33 cm\),最低一级宽\(110 cm\),中间还有\(10\)级,各级宽度依次成等差数列,计算中间各级的宽度.
解析 设梯子的第\(n\)级的宽为\(a_n cm\),其中最高一级宽为\(a_1 cm\),则数列\(\{a_n \}\)是等差数列.
由题意,得\(a_1=33\),\(a_{12}=110\),\(n=12\),
则\(a_{12}=a_1+11d\).
所以\(110=33+11d\),解得\(d=7\).
所以\(a_2=33+7=40\),\(a_3=40+7=47\),…,\(a_{11}=96+7=103\),
即梯子中间各级的宽度从上到下依次是\(40 cm\),\(47 cm\),\(54 cm\),\(61 cm\),\(68 cm\),
\(75 cm\),\(82 cm\),\(89 cm\),\(96 cm\),\(103 cm\).
【巩固练习】
1.《周髀算经》中一个问题:从冬至之日起,小寒、大寒、立春、雨水、惊蛰、春分、清明、谷雨、立夏、小满、芒种这十二个节气的日影子长依次成等差数列,若冬至、立春、春分的日影子长的和是\(37.5\)尺,芒种的日影子长为\(4.5\)尺,则冬至的日影子长为( )
A.\(15.5\)尺 \(\qquad \qquad \qquad \qquad\) B.\(12.5\)尺 \(\qquad \qquad \qquad \qquad\) C.\(10.5\)尺 \(\qquad \qquad \qquad \qquad\) D.\(9.5\)尺
2.《莱茵德纸草书》是世界上最古老的数学著作之一,书中有这样的一道题:把\(120\)个面包分成\(5\)份,使每份的面包数成等差数列,且较多的三份之和恰好是较少的两份之和的\(7\)倍,若将这\(5\)份面包数按由少到多的顺序排列,则第\(4\)份面包的数量为( )
A.\(15\) \(\qquad \qquad \qquad \qquad\) B.\(25\) \(\qquad \qquad \qquad \qquad\) C.\(35\) \(\qquad \qquad \qquad \qquad\) D.\(45\)
3.有一正四棱台形楼顶,其中一个侧面中最上面一行铺瓦\(30\)块,总共需要铺瓦\(15\)行,并且下一行比其上一行多铺\(3\)块瓦,求该侧面最下面一行需铺瓦多少块?
参考答案
-
答案 \(A\)
解析 由题意,从冬至之日起,小寒、大寒、立春、雨水、惊蛰、春分、清明、谷雨、立夏、小满、芒种这十二个节气的日影子长依次成等差数列\(\{a_n \}\),冬至、立春、春分的日影子长的和是\(37.5\)尺,芒种的日影子长为\(4.5\)尺,
所以 \(\left\{\begin{array}{l} a_1+a_4+a_7=3 a_1+9 d=37.5 \\ a_{12}=a_1+11 d=4.5 \end{array}\right.\),解得\(d=-1\),\(a_1=15.5\).
所以冬至的日影子长为\(15.5\)尺.
故选:\(A\). -
答案 \(C\)
解析 设等差数列的公差为\(d\),五份分别设为\(x-2d\),\(x-d\),\(x\),\(x+d\),\(x+2d\),
则\(\left\{\begin{array}{l} x-2 d+x-d+x+x+d+x+2 d=240 \\ 3 x+3 d=7(2 x-3 d) \end{array}\right.\),
解得\(x=24\),\(d=11\),
\(\therefore\)第\(4\)份面包的数量为\(x+d=35\).
故选:\(C\). -
答案 \(72\)
解析 设从上面开始第\(n\)行铺瓦\(a_n\)块,
则数列\(\{a_n \}\)是首项为\(30\),公差为\(3\)的等差数列.
则\(a_{15}=a_1+14d=30+14×3=72\),
即该侧面最下面一行应铺瓦\(72\)块.
分层练习
【A组---基础题】
1.等差数列\(-3,1,5,…\)的第\(15\)项为( )
A.\(40\) \(\qquad \qquad \qquad \qquad\) B.\(53\) \(\qquad \qquad \qquad \qquad\) C.\(63\) \(\qquad \qquad \qquad \qquad\)D.\(76\)
2.下列通项公式表示的数列为等差数列的是( )
A. \(a_n=\dfrac{n}{n+1}\) \(\qquad \qquad \qquad\) B.\(a_n=n^2-1\) \(\qquad \qquad \qquad\) C.\(a_n=5^n\) \(\qquad \qquad \qquad\) D.\(a_n=3n-1\)
3.已知\(\{a_n \}\)是公差为3的等差数列,\(\{b_n\}\)是公差为\(4\)的等差数列,且\(b_n\in N^*\),则\(\{a_{b_n}\}\)为( )
A.公差为\(7\)的等差数列 \(\qquad \qquad \qquad \qquad\) B.公差为\(12\)的等差数列
C.公比为\(12\)的等比数列 \(\qquad \qquad \qquad \qquad\) D.公比为\(81\)的等比数列
4.已知 \(\left\{\dfrac{1}{a_n+1}\right\}\)是等差数列,且\(a_1=\dfrac{1}{4}\),\(a_4=1\),则\(a_11=\)( )
A.\(-12\) \(\qquad \qquad \qquad \qquad\) B.\(-11\) \(\qquad \qquad \qquad \qquad\) C.\(-6\) \(\qquad \qquad \qquad \qquad\) D.\(-5\)
5.(多选)已知数列\(\left\{\dfrac{a_n}{n+2^n}\right\}\)是首项为\(1\),公差为\(d\)的等差数列,则下列判断正确的是( )
A.\(a_1=3\) \(\qquad \qquad \qquad \qquad \qquad \qquad\) B.若\(d=1\),则\(a_n=n^2+2^n\)
C.\(a_2\)可能为\(6\) \(\qquad \qquad \qquad \qquad \qquad\) D.\(a_1,a_2,a_3\)可能成等差数列
6.已知等差数列\(\{a_n \}\)的各项均为正整数,且\(a_8=2021\),则\(a_1\)的最小值是\(\underline{\quad \quad}\).
7.《九章算术》“竹九节”问题中指出,若有一根九节的竹子,自上而下各节的容积成等差数列,上\(5\)节的容积为\(4\)升,下\(4\)节的容积为\(5\)升,问第五节的容积是多少升?
8.已知数列\(\{a_n \}\)的通项公式是\(a_n=7^{n+2}\),求证:数列\(\{\lga_n \}\)是等差数列.
9.已知数列\(\{a_n \}\)满足:\(a_1=2\), \(a_n=2-\dfrac{9}{a_{n-1}+4}(n>1)\),记 \(b_n=\dfrac{1}{a_n+1}\).
(1)求证:数列\(\{b_n\}\)等差数列;\(\qquad \qquad\) (2)求\(a_n\).
10.各项不为\(0\)的数列\(\{a_n \}\)满足 \(\dfrac{a_n}{a_{n-1}}=\dfrac{1}{3 a_{n-1}+1}\left(n \geq 2, \quad n \in N^*\right)\),且\(a_2=-1\).
(1)求证:数列 \(\left\{\dfrac{1}{a_n}\right\}\)为等差数列;
(2)若 \(\dfrac{a_{n+1}}{a_n} \geq \lambda\)对任意\(n\in N^*\)恒成立,求实数\(λ\)的取值范围.
参考答案
-
答案 \(B\)
解析 \(a_1=-3\),\(d=1-(-3)=4\),
故\(a_{15}=a_1+(15-1)d=-3+14×4=53\). -
答案 \(D\)
解析 \(\because\)等差数列的通项是关于\(n\)的一次函数,在四个选项中,只有\(D\)是关\(n\)的一次函数,
\(\therefore\)所给的四个通项中只有\(D\)表示等差数列,
故选:\(D\). -
答案 \(B\)
解析 \(\because\)\(\{a_n \}\)是公差为\(3\)的等差数列,\(\{b_n\}\)是公差为\(4\)的等差数列,且\(b_n\in N^*\),
\(\therefore a_n=a_1+3(n-1)\),\(b_n=b_1+4(n-1)\),
则\(a_{b_n}=a_1+3(b_n-1)=a_1+3[b_1+4(n-1)]=a_1+3b_1+12(n-1)\).
\(\therefore \{a_{b_n}\}\)为公差为\(12\)的等差数列.
故选:\(B\). -
答案 \(C\)
解析 设等差数列 \(\left\{\dfrac{1}{a_n+1}\right\}\)的公差为d,
\(\because \dfrac{1}{a_1+1}=\dfrac{4}{5}\), \(\dfrac{1}{a_4+1}=\dfrac{1}{2}\). \(\therefore \dfrac{1}{2}=\dfrac{4}{5}+3 d\),解得 \(d=-\dfrac{1}{10}\).
\(\therefore \dfrac{1}{a_{11}+1}=\dfrac{4}{5}+10 \times\left(-\dfrac{1}{10}\right)=-\dfrac{1}{5}\),\(\therefore a_{11}=6\).
故选:\(C\). -
答案 \(ACD\)
解析 由已知可得数列\(\left\{\dfrac{a_n}{n+2^n}\right\}\)的通项公式为 \(\dfrac{a_n}{n+2^n}=1+(n-1) d\),
当\(n=1\)时, \(\dfrac{a_1}{1+2}=1\),解得\(a_1=3\),故\(A\)正确;
若\(d=1\),则 \(\dfrac{a_n}{n+2^n}=1+(n-1) \times 1=n\),
所以\(a_n=n^2+n2^n\),故\(B\)错误;
若\(d=0\),则\(a_n=n+2^n\),\(a_2=6\),故\(C\)正确;
若\(a_1,a_2,a_3\)成等差数列,则\(2a_2=a_1+a_3\),
即\(12+12d=14+22d\),解得\(d=-\dfrac{1}{5}\),
故\(a_1,a_2,a_3\)可能成等差数列,故\(D\)正确.
故选:\(ACD\). -
答案 \(5\)
解析 设等差数列\(\{a_n \}\)的公差为\(d(d\in N^*)\),
由\(a_8=a_1+7d\),得\(a_1=2021-7d\);
由于\(2021=7×288+5\),所以\(a_1=7×288+5-7d=7(288-d)+5\),
即当\(d=288\)时,\(a_1\)有最小值\(5\). -
答案 \(1\)
解析 设自上而下各节的容积构成等差数列\(\{a_n \}\),公差为\(d\),
则 \(\left\{\begin{array}{l} a_1+a_2+a_3+a_4+a_5=4 \\ a_6+a_7+a_8+a_9=5 \end{array}\right.\),
化简得 \(\left\{\begin{array}{l} 5 a_1+10 d=4 \\ 4 a_1+26 d=5 \end{array}\right.\),解得 \(\left\{\begin{array}{l} a_1=0.6 \\ d=0.1 \end{array}\right.\),
故\(a_5=a_1+4d=1\). -
证明 设\(b_n=\lga_n=\lg7^{n+2}=(n+2)\lg7\),
则\(b_{n+1}=[(n+1)+2]\lg7=(n+3)\lg7\),
则\(b_{n+1}-b_n=(n+3)\lg7-(n+2)\lg7=\lg7\).
所以数列\(\{b_n \}\)是等差数列,
即数列\(\{\lga_n \}\)是等差数列. -
答案 (1)略;(2) \(a_n=\dfrac{3}{n}-1\).
解析 (1)证明:由 \(a_n=2-\dfrac{9}{a_{n-1}+4}(n>1)\)得
\(a_n+1=3-\dfrac{9}{a_{n-1}+4}=\dfrac{3 a_{n-1}+3}{a_{n-1}+4}(n>1)\),
即 \(\dfrac{1}{a_n+1}=\dfrac{1}{a_{n-1}+1}+\dfrac{1}{3}(n>1) \Rightarrow \dfrac{1}{a_n+1}-\dfrac{1}{a_{n-1}+1}=\dfrac{1}{3}(n>1)\),
而 \(b_n=\dfrac{1}{a_n+1}\),所以 \(b_{n+1}-b_n=\dfrac{1}{3}(n>1)\),
所以数列\(\{b_n\}\)等差数列;
(2)由(1)结合\(a_1=2\),可得\(b_1=\dfrac{1}{3}\),
所以\(b_n=b_1+(n-1)d=\dfrac{1}{3}+\dfrac{1}{3}{n-1}=\dfrac{1}{3} n\),
故 \(a_n=\dfrac{1}{b_n}-1=\dfrac{3}{n}-1\). -
答案 (1) 略;(2)\(\left(-∞,-\dfrac{1}{2}\right]\).
解析 (1)证明:各项不为\(0\)的数列\(\{a_n \}\)满足\(\dfrac{a_n}{a_{n-1}}=\dfrac{1}{3 a_{n-1}+1}\left(n \geq 2, n \in N^*\right)\),
变为 \(a_n=\dfrac{a_{n-1}}{3 a_{n-1}+1}\),
两边取倒数:可得 \(\dfrac{1}{a_n}=\dfrac{1}{a_{n-1}}+3\),即\(\dfrac{1}{a_n}-\dfrac{1}{a_{n-1}}=3\),
由\(a_2=-1\), \(\therefore \dfrac{1}{-1}-\dfrac{1}{a_1}=3\),解得\(a_1=-\dfrac{1}{4}\).
\(\therefore\)数列 \(\left\{\dfrac{1}{a_n}\right\}\)为等差数列,公差为\(3\),首项为\(-4\).
(2)解:由(1)可得: \(\dfrac{1}{a_n}=-4+3(n-1)=3 n-7\),
由 \(\dfrac{a_{n+1}}{a_n} \geq \lambda\)对任意\(n\in N^*\)恒成立, \(\therefore \lambda \leq \dfrac{3 n-7}{3 n-4}\)的最小值.
令 \(f(n)=\dfrac{3 n-7}{3 n-4}=\dfrac{3 n-4-3}{3 n-4}=1-\dfrac{3}{3 n-4}\),
\(n=1\)时,\(f(1)=4\);\(n=2\)时,\(f(2)=-\dfrac{1}{2}\);
\(n≥3\)时,\(f(n)\)单调递增,\(n→+∞\)时,\(f(n)→1\).
\(\therefore λ≤-\dfrac{1}{2}\),
\(\therefore\)实数\(λ\)的取值范围是\(\left(-∞,-\dfrac{1}{2}\right]\).
【B组---提高题】
1.(多选)已知无穷等差数列\(\{a_n \}\)的公差\(d\in N^*\),且\(5,17,23\)是\(\{a_n \}\)中的三项,则下列结论正确的是( )
A.\(d\)的最大值是\(6\) \(\qquad \qquad \qquad \qquad\) B.\(2a_2≤a_8\)
C.\(a_n\)一定是奇数 \(\qquad \qquad \qquad \qquad\) D.\(137\)一定是数列\(\{a_n \}\)中的项
2.在\(△ABC\)中,角\(A\),\(B\),\(C\)的对边分别为\(a\),\(b\),\(c\),角\(A\),\(B\),\(C\)成等差数列,则 \(\dfrac{a+c}{b}\)的取值范围是\(\underline{\quad \quad}\) .
3.数列\(\{a_n \}\)中,\(a_1=1\),\(a_n=3a_{n-1}+3^n-1(n\in N^*,n≥2)\),若存在实数\(λ\),使得数列 \(\left\{\dfrac{a_n+\lambda}{3^n}\right\}\)为等差数列,则\(λ=\)\(\underline{\quad \quad}\).
4.已知数列\(\{a_n \}\)满足\(a_{n+1}=\dfrac{1+a_n}{3-a_n}\left(n \in \boldsymbol{N}^*\right)\),且\(a_1=0\).
(1)求\(a_2\),\(a_3\)的值;
(2)是否存在一个实常数\(λ\),使得数列\(\left\{\dfrac{1}{a_n-\lambda}\right\}\)为等差数列,请说明理由.
参考答案
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答案 \(ABD\)
解析 \(\because\)无穷等差数列\(\{a_n \}\)的公差\(d\in N^*\),且\(5,17,23\)是\(\{a_n \}\)中的三项,
\(\therefore\)设\(\left\{\begin{array}{l} 17-5=12=m d \\ 23-17=6=n d \end{array}\right.\),解得\(d=\dfrac{6}{m-n}\),
\(\therefore d\)的最大值为\(6\),故\(A\)正确;
\(\because a_1≤5,d\in N^*\),
\(\therefore 2a_2-a_8=a_1-5d≤0\),故\(B\)正确;
\(\because d=\dfrac{6}{m-n}\),
\(\therefore\) 当\(m-n=2\)时,\(d=3\),数列可能为\(5,8,11,14,17,20,23,…\),故\(C\)错误;
\(\because 137=23+19×6\),
\(\therefore 137\)一定是等差数列\(\{a_n \}\)中的项,故\(D\)正确.
故选:\(ABD\). -
答案 \((1,2]\)
解析 由角\(A\),\(B\),\(C\)成等差数列,得\(2B=A+C\),
又\(A+B+C=π\),得\(3B=π\),故 \(B=\dfrac{\pi}{3}\),
所以 \(\dfrac{a+c}{b}=\dfrac{\sin A+\sin C}{\sin B}=\dfrac{2}{\sqrt{3}}(\sin A+\sin C)\),
记 \(y=\sin A+\sin C\),又 \(A+C=\dfrac{2 \pi}{3}\),
所以 \(y=\sin A+\sin \left(\dfrac{2 \pi}{3}-A\right)=\sin A+\dfrac{\sqrt{3}}{2} \cos A+\dfrac{1}{2} \sin A\)
\(=\dfrac{3}{2} \sin A+\dfrac{\sqrt{3}}{2} \sin A=\sqrt{3} \sin \left(A+\dfrac{\pi}{6}\right)\),
由 \(A \in\left(0, \dfrac{2 \pi}{3}\right)\),得 \(A+\dfrac{\pi}{6} \in\left(\dfrac{\pi}{6}, \dfrac{5 \pi}{6}\right)\),
故 \(\sin \left(A+\dfrac{\pi}{6}\right) \in\left(\dfrac{1}{2}, 1\right]\),
所以 \(y=\sqrt{3} \sin \left(A+\dfrac{\pi}{6}\right) \in\left(\dfrac{\sqrt{3}}{2}, \sqrt{3}\right]\),
所以 \(\dfrac{a+c}{b} \in(1,2]\). -
答案 \(-\dfrac{1}{2}\)
解析 \(\because a_1=1\),\(a_n=3a_{n-1}+3^n-1(n\in N^*,n≥2)\),
\(\therefore a_n-\dfrac{1}{2}=3(a_{n-1}-\dfrac{1}{2})+3^n\),
两边同时除以\(3^n\)可得, \(\dfrac{a_n-\dfrac{1}{2}}{3^n}=\dfrac{a_{n-1}-\dfrac{1}{2}}{3^{n-1}}+1\).
\(\therefore\) 数列 \(\left\{\dfrac{a_n-\dfrac{1}{2}}{3^n}\right\}\)是等差数列,由题意可得,\(λ=-\dfrac{1}{2}\). -
答案 (1)\(a_2=\dfrac{1}{3}\),\(a_3=\dfrac{1}{2}\); (2) 存在一个实常数\(λ=1\),使得数列\(\left\{\dfrac{1}{a_n-\lambda}\right\}\)是等差数列.
解析 (1)\(a_2=\dfrac{1}{3}\),\(a_3=\dfrac{1}{2}\);
(2)假设存在一个实常数\(λ\),使得数列\(\left\{\dfrac{1}{a_n-\lambda}\right\}\)为等差数列,
则\(\dfrac{1}{a_1-\lambda}, \dfrac{1}{a_2-\lambda}, \dfrac{1}{a_3-\lambda}\)成等差数列,
所以\(\dfrac{2}{a_2-\lambda}=\dfrac{1}{a_1-\lambda}+\dfrac{1}{a_3-\lambda}\),
所以\(\dfrac{2}{\dfrac{1}{3}-\lambda}=\dfrac{1}{0-\lambda}+\dfrac{1}{\dfrac{1}{2}-\lambda}\),解之得\(λ=1\).
因为\(\dfrac{1}{a_{n+1}-1}-\dfrac{1}{a_n-1}=\dfrac{1}{\dfrac{1+a_n}{3-a_n}-1}-\dfrac{1}{a_n-1}\)\(=\dfrac{3-a_n}{2\left(a_n-1\right)}-\dfrac{1}{a_n-1}=\dfrac{1-a_n}{2\left(a_n-1\right)}=-\dfrac{1}{2}\),
又\(\dfrac{1}{a_1-1}=-1\),
所以存在一个实常数\(λ=1\),使得数列 \(\left\{\dfrac{1}{a_n-\lambda}\right\}\)是首项为\(-1\),公差为\(-\dfrac{1}{2}\)的等差数列.
【C组---拓展题】
1.已知等差数列\(\{a_n \}\)中,\(a_1^2+a_6^2=8\),则\(a_2+a_4\)的取值范围是( )
A. \(\left[-\dfrac{2}{5}, \dfrac{2}{5}\right]\) \(\qquad \qquad\) B. \(\left[-\dfrac{\sqrt{13}}{5}, \dfrac{\sqrt{13}}{5}\right]\) \(\qquad \qquad\) C. \(\left[-\dfrac{2 \sqrt{26}}{5}, \dfrac{2 \sqrt{26}}{5}\right]\) \(\qquad \qquad\) D. \(\left[-\dfrac{4 \sqrt{26}}{5}, \dfrac{4 \sqrt{26}}{5}\right]\)
2.已知数列\(\{a_n \}\)的前\(n\)项和为\(S_n\),\(a_1=1\),\(a_n≠0\),\(a_n a_{n+1}=λS_n-1\),其中\(λ\)为常数.
(1)证明:\(a_{n+2}-a_n=λ\);
(2)是否存在λ,使得\(\{a_n \}\)为等差数列?并说明理由.
参考答案
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答案 \(D\)
解析 设等差数列\(\{a_n \}\)的公差为\(d\)\(a_6=a_1+5d\),
\(\because a_1^2+a_6^2=8,\therefore a_1^2+(a_1+5d)^2=8\),
故令\(a_1=2 \sqrt{2} \cos \theta\), \(a_1+5 d=2 \sqrt{2} \sin \theta\),
则\(d=\dfrac{1}{5}(2 \sqrt{2} \sin \theta-2 \sqrt{2} \cos \theta)\),
\(a_2+a_4=2 a_1+4 d=4 \sqrt{2} \cos \theta+\dfrac{4}{5}(2 \sqrt{2} \sin \theta-2 \sqrt{2} \cos \theta)\)\(=2 \sqrt{2}\left(\dfrac{6}{5} \cos \theta+\dfrac{4}{5} \sin \theta\right)\),
且\(2 \sqrt{2} \cdot \sqrt{\left(\dfrac{6}{5}\right)^2+\left(\dfrac{4}{5}\right)^2}=\dfrac{4 \sqrt{26}}{5}\),
故\(-\dfrac{4 \sqrt{26}}{5} \leq a_2+a_4 \leq \dfrac{4 \sqrt{26}}{5}\),
故选:\(D\). -
答案 (1)略;(2) 存在\(λ=4\)使得数列\(\{a_n \}\)为等差数列.
解析 (1)证明:\(\because a_n a_{n+1}=λS_n-1\),
\(\therefore a_{n+1} a_{n+2}=λS_{n+1}-1\),
两式相减可得\(a_{n+1} (a_{n+2}-a_n)=λa_{n+1}\),
\(\because a_{n+1}≠0\),\(\therefore a_{n+2}-a_n=λ\).
(2)解:\(\because a_n a_{n+1}=λS_n-1\) , \(\therefore a_1 a_2=λS_1-1⇒a_2=λ-1\)
\(\because a_{n+2}-a_n=λ\) ,\(\therefore a_3=a_1+λ=λ+1\)
假设存在\(λ\),使得\(\{a_n \}\)为等差数列,则\(a_1,a_2,a_3\)成等差数列,
即\(2a_2=a_1+a_3\),
\(\therefore 2(λ-1)=1+λ+1\),解得\(λ=4\),
故\(a_{n+2}-a_n=4\),
可知数列\(\{a_n \}\)中偶数项可组成首项为\(a_2=3\),公差为\(4\)的等差数列,
即\(a_{2n}=3+4(n-1)=4n-1=2\cdot (2n)-1\);
数列\(\{a_n \}\)中奇数项可组成首项为\(a_1=1\),公差为\(4\)的等差数列,
即 \(a_{2 n-1}=1+4(n-1)=4 n-3=2 \cdot(2 n-1)-1\);
所以\(a_n=2n-1\),
则\(a_{n+1}-a_n=2\),
因此存在\(λ=4\)使得数列\(\{a_n \}\)为等差数列.