4.2 等差数列及其前n项和
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知识剖析
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)
\({\color{Red}{解释}}\)
\(a_n-a_{n-1}=2\)\((n \geq 2)\)\(⇒\left\{a_{n}\right\}\)是公差为\(2\)的等差数列;
\(a_{n+1}-a_n=-3\)\(⇒\left\{a_{n}\right\}\)是公差为\(-3\)的等差数列;
\(a_{n+1}-a_n=3n\)\(⇒\left\{a_{n}\right\}\)不是等差数列.
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\(b=\dfrac{a+c}{2}\).
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)
前n项和
等差数列\(\{a_n\}\)的首项为\(a_1\),公差为\(d\),则其前\(n\)项和为
\(S_{n}=\dfrac{\left(a_{1}+a_{n}\right) n}{2}\)(由倒序相加法可证)
\(S_{n}=n a_{1}+\dfrac{n(n-1)}{2} d\)
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,\(n∈N^*\))\(⇒a_n\)是等差数列;
② 中项法:\(2 a_{n+1}=a_{n}+a_{n+2}\)\((n∈N^*)\)\(⇒a_n\)是等差数列;
③ 通项公式法:\(a_n=kn+b\)(\(k ,b\)是常数)\(⇒a_n\)是等差数列;
④ 前\(n\)项和公式法:\(S_n=A n^2+Bn\)(\(A ,B\)是常数)\(⇒a_n\)是等差数列;
\({\color{Red}{PS}}\) 方法③④不可以在解答题里直接使用.
基本性质
(其中\(m ,n ,p ,t∈N^*\))
若数列\(\{a_n\}\)是首项为\(a_1\),公差为\(d\)的等差数列,它具有以下性质:
\((1)\)若\(m+n=p+t\), 则\(a_m+a_n=a_p+a_t\);
\((2)\)\(a_n=a_m+(n-m) d\);
\((3)\)\(d=\dfrac{a_{n}-a_{m}}{n-m}\);
\((4)\)下标成等差数列且公差为\(m\)的项\(a_{k}\),\(a_{k+m}\),\(a_{k+2 m}\),…(\(k ,m∈N^*\))组成公差为\(md\)的等差数列;
\((5)\)数列\(\{λa_n+b\}\)(\(λ、b\)是常数)是公差是\(λb\)的等差数列;
\((6)\)若数列\(\{b_n\}\)也是等差数列,则数列\(\{a_n±b_n\}\),\(\{ka_n±b_n\}\)(\(k\)为非零常数)也是等差数列;
\((7)\)\(S_n\),\(S_{2n}-S_n\),\(S_{3n}-S_{2n}\)…成等差数列;
\((8)\)\(S_{2n-1}=(2n-1) a_n\).
经典例题
【题型一】等差数列的基本运算
【典题1】已知\(\{a_n\}\)为等差数列,若\(a_3=1\),\(S_4=0\),则\(a_6=\) \(\underline{\quad \quad}\).
【解析】
\({\color{Red}{(a_3,a_6用到通项公式a_n=a_1+(n-1)d,S_4用前n项和公式S_{n}=n a_{1}+\dfrac{n(n-1)}{2} d)}}\)
设等差数列\(\{a_n\}\)的公差为\(d\),
由\(\left\{\begin{array}{l}
a_{3}=1 \\
S_{4}=0
\end{array},\right.\),得\(\left\{\begin{array}{l}
a_{1}+2 d=1 \\
4 a_{1}+\dfrac{4 \times 3}{2} d=0
\end{array}\right.\),
\({\color{Red}{(得到a_1,d的方程组)}}\)
解得\(\left\{\begin{array}{l}
a_{1}=-3 \\
d=2
\end{array}\right.\),所以\(a_6=a_1+5d=7\).
【点拨】
①首项\(a_1\)、公差\(d\)是等差数列的“基本量”,若知道它们数列其他量就可求.故提示我们,题中\(a_m\)用上通项公式\(a_n=a_1+(n-1)d\),对\(S_m\)用上前\(n\)项和\(S_{n}=n a_{1}+\dfrac{n(n-1)}{2} d\).
② 若已知\(a_1\),\(d\),\(n\),\(a_n\),\(S_n\)中三个便可求出其余两个,即“知三求二”,实质是利用方程思想建立方程组进行求解.
【典题2】已知等差数列\(\{a_n\}\)的公差不为\(0\),其前\(n\)项和为\(S_n\),且\(2a_1\),\(S_8\),\(S_9\)成等差数列,则下列四个选项中正确的有( )
A.\(2a_5+3a_9=S_8\) \(\qquad \qquad \qquad \qquad\) B.\(S_2=S_7\) \(\qquad \qquad \qquad \qquad\) C.\(a_5=0\) \(\qquad \qquad \qquad \qquad\) D.\(S_5\)最小
【解析】设等差数列的公差为\(d\),
则\(S_{8}=8 a_{1}+\dfrac{8 \times 7}{2} d=8 a_{1}+28 d\),
\(S_{9}=9 a_{1}+\dfrac{9 \times 8}{2} d=9 a_{1}+36 d\)
\({\color{Red}{(把S_8、S_9化为关于a_1、d的式子)}}\)
由题意可知:\(2S_8=2a_1+S_9\),
即\(16a_1+56d=2a_1+9a_1+36d\),
解得\(a_1=-4d\),
\({\color{Red}{(得到a_1、d的关系)}}\)
\(\therefore a_{n}=a_{1}+(n-1) d=(n-5) d\),\(S_{n}=n a_{1}+\dfrac{n(n-1)}{2} d=\dfrac{\left(n^{2}-9 n\right) d}{2}\)
对于\(A\)选项,\(2a_5+3a_9=3×4d=12d\),\(S_{8}=\dfrac{\left(8^{2}-8 \times 9\right) d}{2}=-4 d\),\(A\)项错误,
对于\(B\)选项,\(S_{2}=\dfrac{\left(2^{2}-9 \times 2\right) d}{2}=-7 d\),\(S_{7}=\dfrac{\left(7^{2}-9 \times 7\right) d}{2}=-7 d\),\(B\)选项正确,
对于\(C\)选项,\(a_5=0\),\(C\)选项正确.
对于\(D\)选项,
\({\color{Red}{方法一}}\)\(S_{n}=\dfrac{d}{2}\left(n^{2}-9 n\right)=\dfrac{d}{2}\left[\left(n-\dfrac{9}{2}\right)^{2}-\dfrac{81}{4}\right]\),\({\color{Red}{(利用二次函数性质)}}\)
若\(d>0\),则\(S_4\)或\(S_5\)最小;若\(d<0\),则\(S_4\)或\(S_5\)最大.\(D\)选项错误.
\({\color{Red}{方法二}}\)由\(a_n=(n-5)d\)可知,
若\(d>0\),则数列\(\{a_n\}\)是增函数,且\(a_5=0\),从第\(6\)项开始为正数,即\(S_4\)或\(S_5\)最小;
若\(d<0\),则数列\(\{a_n\}\)是减函数,且\(a_5=0\),从第\(6\)项开始为负数,即\(S_4\)或\(S_5\)最大.
故选:\(BC\).
【点拨】
① 本题充分利用到了方程思想和数列的基本量,把题中的\(a_m\)用通项公式\(a_n=a_1+(n-1)d\)表示,\(S_m\)用前\(n\)项和\(S_{n}=n a_{1}+\dfrac{n(n-1)}{2} d\)表示,都转化为基本量.
② 求等差数列前\(n\)项和的最值的方法
(1)求出\(S_n\),再利用二次函数的性质;
(2)求出\(a_n\),知晓数列的单调性判断前\(n\)项和是求最小值还是最大值,求使得\(a_n≥0\)(或\(a_n≤0\))成立时最大的\(n\)值.
【典题3】设\(d\)为正项等差数列\(\{a_n\}\)的公差,若\(d>0\),\(a_3=2\),则( )
A.\(a_2 a_4<4\) \(\qquad \qquad\) B.\(a_{2}^{2}+a_{4} \geq \dfrac{15}{4}\) \(\qquad \qquad\)C.\(a_1 a_5>a_2 a_4\) \(\qquad \qquad\)D.\(\dfrac{1}{a_{1}}+\dfrac{1}{a_{5}}>1\)
【解析】由题设知\(\left\{\begin{array}{l}
d>0 \\
a_{1}=2-2 d>0
\end{array}\right.\),
解得:\(0<d<1\),\({\color{Red}{(得到d的范围)}}\)
对于\(A\)
\(∵a_2 a_4=(2-d)(2+d)=4-d^2<4\),\(∴A\)正确;
对于\(B\)
\(\because a_{2}^{2}+a_{4}=(2-d)^{2}+2+d=d^{2}-3 d+6>4>\dfrac{15}{4}\),\(∴B\)正确;
对于\(C\)
\(∵a_1 a_5-a_2 a_4=(2-2d)(2+2d)-(2-d)(2+d)=-3d^2<0\),
\(∴a_1 a_5<a_2 a_4\),\(C\)错误.
\({\color{Red}{(ABC中均是把涉及的量转化为d的式子进行判断)}}\)
对于\(D\)
\(∵a_1+a_5=2a_1+4d=4\),
\(\therefore \dfrac{1}{a_{1}}+\dfrac{1}{a_{5}}=\dfrac{1}{4}\left(\dfrac{1}{a_{1}}+\dfrac{1}{a_{5}}\right)\left(a_{1}+a_{5}\right)\)\(=\dfrac{1}{4}\left(2+\dfrac{a_{5}}{a_{1}}+\dfrac{a_{1}}{a_{5}}\right)>\dfrac{1}{4}\left(2+2 \sqrt{\dfrac{a_{5}}{a_{1}} \cdot \dfrac{a_{1}}{a_{5}}}\right)=1\)
\({\color{Red}{(此处由于d>0,a_1不可能等于a_5,则取不到等号)}}\)
\(∴D\)正确;故选:\(ABD\).
【点拨】对于不等式的处理,也可以使用基本量\(a_1\)、\(d\)的方法求解.
巩固练习
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\(a_5=\) \(\underline{\quad \quad}\) .
2(★)等差数列\(\{a_n\}\)满足\(a_2+a_5=3\),\(a_6=2\),则\(a_4+a_7=\) \(\underline{\quad \quad}\).
3(★)【多选题】记\(S_n\)为等差数列\(\{a_n\}\)的前\(n\)项和,若\(a_4+a_5=24\),\(S_6=48\),则下列正确的是( )
A.\(a_1=-2\) \(\qquad \qquad \qquad \qquad\) B.\(a_1=2\)\(\qquad \qquad \qquad \qquad\) C.\(d=4\) \(\qquad \qquad \qquad \qquad\) D.\(d=-4\)
4(★★)【多选题】设数列\(\{a_n\}\)是等差数列,\(S_n\)是其前\(n\)项和,\(a_1>0\)且\(S_6=S_9\),则( )
A.\(d>0\) \(\qquad \qquad\) B.\(a_8=0\) \(\qquad \qquad\) C.\(S_7\)或\(S_8\)为\(S_n\)的最大值 \(\qquad \qquad\) D.\(S_5>S_6\)
5(★★)【多选题】等差数列\(\{a_n\}\)的前\(n\)项和为\(S_n\),\(a_1+5a_3=S_8\),则下列结论一定正确的是( )
A.\(a_{10}=0\)
B.当\(n=9\)或\(10\)时,\(S_n\)取最大值
C.\(|a_9 |<|a_{11}|\)
D.\(S_6=S_{13}\)
6(★★)【多选题】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.\(a_1=3\)
B.若\(d=1\),则\(a_n=n^2+2^n\)
C.\(a_2\)可能为\(6\)
D.\(a_1 ,a_2 ,a_3\)可能成等差数列
7(★★)【多选题】已知无穷等差数列\(\{a_n\}\)的公差\(d∈N^*\),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
B.\(2a_2≤a_8\)
C.\(a_n\)一定是奇数
D.\(137\)一定是数列\(\{a_n\}\)中的项
8(★★)已知等差数列\(\{a_n\}\)的前\(n\)项和为\(S_n\),若\(a_n a_{n+1}\),\(2S_n\)\(,1\)成等差数列.
(1)求数列\(\{a_n\}\)的通项公式;
(2)求数列\(\left\{\dfrac{a_{n+1}}{8-a_{n}}\right\}\)的最大项与最小项.
参考答案
- \(\dfrac{4}{3}\)
- \(\dfrac{19}{5}\)
- \(AC\)
- \(BC\)
- \(AD\)
- \(ACD\)
- \(ABD\)
- \((1) a_n=2n-1\)
\((2)\)最大项是第\(4\)项,值为\(9\);最小项是第\(5\)项,值为\(-11\)
【题型二】等差数列的判断与证明
【典题1】数列\(\{a_n\}\)的前\(n\)项和\(S_n=33n-n^2\).
(1)求数列\(\{a_n\}\)的通项公式;
(2)求证:\(\{a_n\}\)是等差数列.
【解析】(1)当\(n≥2\)时,\(a_n=S_n-S_{n-1}=34-2n\)
又当\(n=1\)时,\(a_1=S_1=32\)满足上式,
故\(\{a_n\}\)的通项为\(a_n=34-2n\).
(2)证明:\(a_{n+1}-a_n=34-2(n+1)-(34-2n)=-2\).
故数列\(\{a_n\}\)是以\(32\)为首项,\(-2\)为公差的等差数列.
【点拨】
① 证明数列方法有定义法:\(a_{n+1}-a_n=d\)(\(d\)是常数,\(n∈N^*\))
② 通过等差数列的通项公式和前\(n\)项和公式,也可知道以下两种方法,但在解答题中不能直接使用,
通项公式法:\(a_n=kn+b\)(\(k ,b\)是常数)\(⟹a_n\)是公差为\(k\)等差数列;
前\(n\)项和公式法:\(S_n=An^2+Bn\)(\(A ,B\)是常数)\(⟹a_n\)是等差数列.
【典题2】已知数列\(\{a_n\}\)满足:\(a_1=2\),\(a_{n}=2-\dfrac{9}{a_{n-1}+4}\)\((n>1)\),
(1)求证:数列\(\left\{\dfrac{1}{a_{n}+1}\right\}\)等差数列;
(2)求\(a_n\).
【解析】(1)证明:记\(b_{n}=\dfrac{1}{a_{n}+1}\).
由\(a_{n}=2-\dfrac{9}{a_{n-1}+4}\)\((n>1)\)得\(a_{n+1}=2-\dfrac{9}{a_{n}+4}\)
则\(b_{n+1}-b_{n}=\dfrac{1}{a_{n+1}+1}-\dfrac{1}{a_{n}+1}\)\(=\dfrac{1}{2-\dfrac{9}{a_{n}+4}+1}-\dfrac{1}{a_{n}+1}\)\(=\dfrac{a_{n}+4}{3 a_{n}+3}-\dfrac{1}{a_{n}+1}=\dfrac{a_{n}+1}{3\left(a_{n}+1\right)}=\dfrac{1}{3}\)
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\(a_{n}=\dfrac{1}{b_{n}}-1=\dfrac{3}{n}-1\)
【点拨】
① 本题是由递推公式求通项公式的题型,题目先求证\(\left\{\dfrac{1}{a_{n}+1}\right\}\)是等差数列,避免构造新数列,从而降低了难度;
② 利用等差数列的定义法,只需要求证\(b_{n+1}-b_n\)是常数,过程仅仅需要运算化简,没太多的技巧要求.
【典题3】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.
【解析】(1)证明:\(∵a_n a_{n+1}=λS_n-1\),
\({\color{Red}{(已知条件是a_n与S_n的关系式,易想到a_{n}=\begin{cases}S_{1},& n=1 \\S_{n}-S_{n-1},& n\geq 2\end{cases})}}\)
\(\therefore a_{n+1} a_{n+2}=\lambda S_{n+1}-1\),
两式相减可得\(a_{n+1}\left(a_{n+2}-a_{n}\right)=\lambda a_{n+1}\)
\(\because a_{n+1} \neq 0\),\(\therefore a_{n+2}-a_{n}=\lambda\).
(2)解:\(\because a_{n} a_{n+1}=\lambda S_{n}-1\),\(a_{1}=1\)
\(\therefore a_{1} a_{2}=\lambda S_{1}-1 \Rightarrow a_{2}=\lambda-1\)
\(\because a_{n+2}-a_{n}=\lambda\)
\(\therefore a_{3}=a_{1}+\lambda=\lambda+1\)
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\(2a_2=a_1+a_3\),
\({\color{Red}{(先通过前3项成等差数列证明\{a_n\}为等差数列的必要条件)}}\)
\(∴2(λ-1)=1+λ+1\),解得\(λ=4\),
\({\color{Red}{(得到λ后还要证明其充分性)}}\)
故\(a_{n+2}-a_n=4\),
\({\color{Red}{(由于a_{n+2}、a_n是隔项,故分奇偶项进行讨论)}}\)
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\(a_{2n-1}=1+4(n-1)=4n-3=2\cdot (2n-1)-1\)
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.
【点拨】本题求证是否存在\(λ\),使得数列\(\{a_n\}\)为等差数列,思路是先利用前\(3\)项成等差数列证明其必要性,再证明其充分性,过程才完整严谨.
巩固练习
1(★)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\(b_n∈N^*\),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
2(★★)设数列\(\{a_n\}\)的前\(n\)项和为\(S_n\),且对任意正整数\(n\),\(a_n+s_n=4096\).若\(b_{n}=\log _{2} a_{n}\),则数列\(\{b_n\}\)为( )
A.公差为\(-1\)的等差数列 \(\qquad \qquad \qquad \qquad\) B.公差为\(1\)的等差数列
C.公比数列为\(\dfrac{1}{2}\)的等比数列 \(\qquad \qquad \qquad \qquad\) D.公比数列为\(-\dfrac{1}{2}\)的等比数列
3(★★)数列\(\{a_n\}\)中,已知\(S_1=1\),\(S_2=2\),且\(S_{n+1}+2 S_{n-1}=3 S_{n}\),(\(n≥2\)且\(n∈N^*\)),则此数列为( )
A.等差数列 \(\qquad \qquad \qquad \qquad \qquad \qquad\) B.等比数列
C.从第二项起为等差数列 \(\qquad \qquad \qquad \qquad\) D.从第二项起为等比数列
4(★★)数列\(\{a_n\}\)中,\(a_1=1\),\(a_n=3a_{n-1}+3^n-1\)\((n∈N^* ,n≥2)\),若存在实数\(λ\),使得数列\(\left\{\dfrac{a_{n}+\lambda}{3^{n}}\right\}\)为等差数列,则\(λ=\) \(\underline{\quad \quad}\) .
5(★★)已知数列\(\{a_n\}\)的前\(n\)项和为\(S_n=2n^2-3n-2\),求证数列\(\left\{\dfrac{S_{n}}{2 n+1}\right\}\)是等差数列.
6(★★)已知数列\(\{a_n\}\)中,\(a_1=4\),\(a_n=a_{n-1}+2^{n-1}+3(n≥2)\).证明数\(\left\{a_{n}-2^{n}\right\}\)是等差数列,并求\(\{a_n\}\)的通项公式.
7(★★★)设数列\(\{a_n\}\)满足:\(a_{1}=\dfrac{1}{2}\),\(a_{n+1}-a_{n}=2\left(a_{n+1}-1\right)\left(a_{n}-1\right)\),证明数列\(\left\{\dfrac{1}{a_{n}-1}\right\}\)是等差数列并求数列\(\{a_n\}\)的通项公式\(a_n\).
8(★★★)已知数列\(\{a_n\}\)满足\(a_{n+1}=\dfrac{1+a_{n}}{3-a_{n}}\)\((n∈N^*)\),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.
答案
1.\(B\)
2.\(A\)
3.\(D\)
4.\(\lambda=-\dfrac{1}{2}\)
5. 提示:作差法
6. 证明略,\(a_n=2^n+3n-1\)
7. 证明略,\(a_{n}=\dfrac{2 n-1}{2 n}\)
8.\(\text { (1) } a_{2}=\dfrac{1}{3}, a_{3}=\dfrac{1}{2}\)
\((2)\)存在一个实常数\(λ=1\),使得数列\(\left\{\dfrac{1}{a_{n}-\lambda}\right\}\)为等差数列.
【题型三】等差数列的基本性质及运用
【典题1】已知等差数列\(\{a_n\}\)满足\(a_1+a_3+a_5=18\),\(a_3+a_5+a_7=30\),则\(a_2+a_4+a_6=\) \(\underline{\quad \quad}\).
【解析】 \({\color{Red}{方法一}}\) 设公差为\(d\),
则\(a_1+a_3+a_5\),\(a_2+a_4+a_6\),\(a_3+a_5+a_7\)
显然也成等差数列,且公差为\(3d\),
\(∴2(a_2+a_4+a_6 )=(a_1+a_3+a_5 )+(a_3+a_5+a_7 )=48\)
\(∴a_2+a_4+a_6=24\),
\({\color{Red}{方法二}}\)\(∵a_1+a_3+a_5=18\)
\(∴3a_3=18⇒a_3=6\),
\({\color{Red}{(等差数列性质:若m+n=p+t, 则a_m+a_n=a_p+a_t)}}\)
\(∵a_3+a_5+a_7=30\),
\(∴3a_5=30⇒a_5=10\)
\(\therefore a_{2}+a_{4}+a_{6}=3 a_{4}=3 \times \dfrac{a_{3}+a_{5}}{2}=24\).
【点拨】本题当然可以用方程思想求解,利用等差数列性质起到降低计算量的效果,这需要善于观察小标之间的关系.
【典题2】已知两个等差数列\(\{a_n\}\),\(\{b_n\}\)的前\(n\)项和分别为\(S_n\),\(T_n\),若对任意的整数\(n\),都有\(\dfrac{S_{n}}{T_{n}}=\dfrac{2 n-7}{3 n+2}\),则\(\dfrac{a_{5}}{b_{1}+b_{11}}+\dfrac{a_{7}}{b_{3}+b_{9}}\)等于\(\underline{\quad \quad}\).
【解析】依题意,数列\(\{a_n\}\),\(\{b_n\}\)均为等差数列,
\(\therefore \dfrac{a_{5}}{b_{1}+b_{11}}+\dfrac{a_{7}}{b_{3}+b_{9}}\)
\(=\dfrac{a_{5}}{2 b_{6}}+\dfrac{a_{7}}{2 b_{6}}=\dfrac{a_{5}+a_{7}}{2 b_{6}}\)
\({\color{Red}{(等差数列性质:若m+n=p+t, 则a_m+a_n=a_p+a_t)}}\)
\(=\dfrac{a_{6}}{b_{6}}=\dfrac{S_{11}}{T_{11}}\)
\({\color{Red}{(等差数列性质:S_{2 n-1}=(2 n-1) a_{n})}}\)
\(=\dfrac{3}{7}\).
【典题3】已知\(S_n\)是等差数列\(\{a_n\}\)的前\(n\)项和,且\(S_6>S_7>S_5\),给出下列五个命题:
①\(d<0\); \(\qquad \qquad\) ②\(S_{11}>0\); \(\qquad \qquad\) ③\(S_{12}<0\);
④数列\(\{S_n\}\)中的最大项为\(S_{11}\); \(\qquad \qquad\) ⑤\(|a_6 |>|a_7 |\)
其中正确的命题是 \(\underline{\quad \quad}\).
【解析】 \({\color{Red}{方法一}}\) \(∵S_6>S_7>S_5\),
\(∴a_6=S_6-S_5>0\),\(a_7=S_7-S_6<0\),\(a_6+a_7=S_7-S_5>0\)
①\(d=a_7-a_6<0\), 所以①正确;
②\(S_{11}=\dfrac{11\left(a_{1}+a_{11}\right)}{2}=11 a_{6}>0\),故②正确;
③\(S_{12}=6(a_1+a_{12})=6(a_6+a_7 )>0\),故③错误;
④\(∵a_6>0 ,a_7<0\),\(∴\)数列\(\{S_n\}\)中的最大项为\(S_6\),故④错误;
⑤\(∵a_6>0 ,a_7<0\),\(a_6+a_7>0\),\(\therefore\left|a_{6}\right|>\left|a_{7}\right|\),故⑤正确.
综上,①②⑤正确.
\({\color{Red}{方法二}}\)等差数列的前\(n\)项和\(S_{n}=a_{1} n+\dfrac{n(n-1)}{2} d=\dfrac{d}{2} n^{2}+\left(a_{1}-\dfrac{d}{2}\right) n\)
由题意可知\(d≠0\),则它是关于\(n\)的二次函数,
由\(S_6>S_7>S_5\),想象下图象,
若图象开口向上,则\(n=6\)比\(n=5\)和\(n=7\)都要离对称轴远,是不可能的;
可得图象开口向下,且对称轴在\((6,6.5)\),\(12<x_0<13\)(\(x_0\)为二次函数零点),
则\(d<0\),\(S_{11}>0\),\(S_{12}>0\),数列\(\{S_n\}\)中的最大项为\(S_6\),\(|a_6 |>|a_7 |\).
综上,①②⑤正确.
【点拨】
① 本题是不等式问题,若使用基本量\(a_1,d\)表示,计算量较大,思路显得呆板,利用性质求解更简洁,也更看清楚其本质;
② 处理其\(n\)项和问题(比如比较大小,求最值等),利用\(S_{n}=\dfrac{d}{2} n^{2}+\left(a_{1}-\dfrac{d}{2}\right) n\)其对应的函数图象较容易得出结果.
巩固练习
1(★)在等差数列\(\{a_n\}\)中,\(a_3+a_8+a_{13}=27\),\(S_n\)表示数列\(\{a_n\}\)的前\(n\)项和,则\(S_{15}=\) \(\underline{\quad \quad}\).
2(★)已知各项不为\(0\)的等差数列\(\{a_n\}\)的前\(n\)项和为\(S_n\),若\(a_5=2a_2\),则\(\dfrac{S_{9}}{a_{2}}=\) \(\underline{\quad \quad}\).
3(★★)两个等差数列\(\{a_n\}\),\(\{b_n\}\)的前\(n\)项和为\(S_n\),\(T_n\),且\(\dfrac{S_{n}}{T_{n}}=\dfrac{2 n-3}{3 n-2}\),则\(\dfrac{a_{5}}{b_{5}}=\) \(\underline{\quad \quad}\).
4(★★)已知等差数列\(\{a_n\}\)满足\(a_1>0\),\(a_{2019}+a_{2020}>0\),\(a_{2019}⋅a_{2020}<0\).其前\(n\)项和为\(S_n\),则使\(S_n>0\)成立时\(n\)最大值为\(\underline{\quad \quad}\).
5(★★)设正项等差数列\(\{a_n\}\)的前\(n\)项和为\(S_n\),若\(S_{21}=63\),则\(\dfrac{1}{a_{5}}+\dfrac{4}{a_{17}}\)的最小值为\(\underline{\quad \quad}\).
6(★★)设等差数列\(\{a_n\}\)的前\(n\)项和为\(S_n\),若\(a_6<0\),\(a_7>0\),且\(a_7>|a_6 |\),则( )
A.\(S_{11}+S_{12}<0\) \(\qquad \qquad\) B.\(S_{11}+S_{12}>0\) \(\qquad \qquad\)C.\(S_{11}S_{12}<0\) \(\qquad \qquad\) D.\(S_{11} S_{12}>0\)
7(★★)已知等差数列\(\{a_n\}\)的前\(n\)项和为\(S_n\),\(S_{4}=40\),\(S_{n}=210\),\(S_{n-4}=130\),则\(n=\)( )
A.\(12\) \(\qquad \qquad \qquad \qquad\)B.\(14\)\(\qquad \qquad \qquad \qquad\) C.\(16\) \(\qquad \qquad \qquad \qquad\)D.\(18\)
8(★★)【多选题】 等差数列\(\{a_n\}\)的前\(n\)项和为\(S_n\),若\(a_1>0\),公差\(d≠0\),则下列命题正确的是( )
A.若\(S_5=S_9\),则必有\(S_{14}=0\)
B.若\(S_5=S_9\),则必有\(S_7\)是\(S_n\)中最大的项
C.若\(S_6>S_7\),则必有\(S_7>S_8\)
D.若\(S_6>S_7\),则必有\(S_5>S_6\)
9(★★) 设正项等差数列\(\{a_n\}\)满足\(\left(a_{1}+a_{10}\right)^{2}=2 a_{2} a_{9}+20\),则( )
A.\(a_2 a_9\)的最大值为\(10\)
B.\(a_2+a_9\)的最大值为\(2 \sqrt{10}\)
C.\(\dfrac{1}{a_{2}^{2}}+\dfrac{1}{a_{9}^{2}}\)的最大值为\(\dfrac{1}{5}\)
D.\(a_2^4+a_9^4\)的最小值为\(200\)
参考答案
1.\(135\)
2.\(18\)
3.\(\dfrac{3}{5}\)
4.\(4038\)
5.\(\dfrac{3}{2}\)
6.\(C\)
7.\(B\)
8.\(ABC\)
9.\(ABD\)