4.2.2 等差数列的综合应用
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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\(d\) .
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: \(2a_{n+1}=a_n+a_{n+2} (n∈N^*)⟹a_n\)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;
注 方法③④不可以在解答题里直接使用.
基本性质
以下内容中\(m\) ,\(n\) ,\(p\) ,\(t∈N^*\).
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) \(S_n\) , \(S_{2 n}-S_n\), \(S_{3 n}-S_{2 n}\)\(…\)\((n∈N^*)\)成等差数列;
(8) \(S_{2 n-1}=(2 n-1) a_n\).
基本方法
【题型1】 等差数列的基本运算
【典题1】 等差数列\(\{a_n \}\)的前\(n\)项和为\(S_n\),已知\(a_1+2a_3=-1\),\(S_4=0\).则\(S_n\)的最小值为( )
A. \(-4\) \(\qquad \qquad \qquad \qquad\) B.\(-3\) \(\qquad \qquad \qquad \qquad\) C. \(-2\) \(\qquad \qquad \qquad \qquad\) D.\(-1\)
解析 设等差数列\(\{a_n \}\)的公差为\(d\),
\(\because a_1+2a_3=-1\),\(S_4=0\),
\(\therefore\left\{\begin{array}{l}
3 a_1+4 d=-1 \\
4 a_1+6 d=0
\end{array}\right.\),解得\(a_1=-3\),\(d=2\),
故\(a_n=-3+(n-1)×2=2n-5\),
当\(n=2\)时,\(a_2=-1<0\),当\(n=3\)时,\(a_3=1\),
故\(S_n\)的最小值为\(S_2=a_1+a_2=-3+(-1)=-4\).
故选:\(A\).
【巩固练习】
1.已知数列\(\{a_n \}\)是等差数列,其前\(n\)项和为\(S_n\),且\(a_1=1\),\(S_8=4S_4\),若\(a_k+a_3=18\),则\(k\)的值为( )
A. \(6\) \(\qquad \qquad \qquad \qquad\) B. \(7\) \(\qquad \qquad \qquad \qquad\) C. \(8\) \(\qquad \qquad \qquad \qquad\) D.\(9\)
2.设等差数列\(\{a_n \}\)的前\(n\)项和为\(S_n\),\(S_{35}<0\),\(S_{36}>0\).若对任意的正整数\(n\),都有\(S_n≥S_k\),则整数\(k=\)( )
A. \(34\) \(\qquad \qquad \qquad \qquad\) B. \(35\) \(\qquad \qquad \qquad \qquad\) C. \(18\) \(\qquad \qquad \qquad \qquad\) D.\(19\)
参考答案
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答案 \(B\)
解析 在等差数列\(\{a_n \}\)中,设公差为\(d\),
由\(a_1=1\),\(S_8=4S_4\),得\(8 \times 1+\dfrac{8 \times 7}{2} d=4\left(4 \times 1+\dfrac{4 \times 3}{2} d\right)\),
解得\(d=2\).
若\(a_k+a_3=18\),则\(1+2(k-1)+1+2×2=18\),解得\(k=7\).
故选:\(B\). -
答案 \(C\)
解析 在等差数列\(\{a_n \}\)中,由\(S_{35}<0\),\(S_{36}>0\),
得 \(\left\{\begin{array}{l} \dfrac{\left(a_1+a_{35}\right) \times 35}{2}<0 \\ \dfrac{\left(a_1+a_{36}\right) \times 36}{2}>0 \end{array}\right.\),则 \(\left\{\begin{array}{l} a_{18}<0 \\ a_{18}+a_{19}>0 \end{array}\right.\),
可得 \(a_{18}<0\), \(a_{19}>0\),且 \(a_{19}>\left|a_{18}\right|\),
若对任意的正整数\(n\),都有\(S_n≥S_k\),则\(k=18\).
故选:\(C\).
【题型2】 等差数列的基本性质及运用
【典题1】 等差数列\(\{a_n \}\)的前\(n\)项和为\(S_n\),若\(S_5<S_6\),\(S_6=S_7\),\(S_7>S_8\),则下列结论错误的是( )
A. \(a_6+a_8=0\) \(\qquad \qquad \qquad\) B. \(S_5=S_8\) \(\qquad \qquad \qquad\) C. 数列\(\{a_n \}\)是递减数列 \(\qquad \qquad \qquad\) D.\(S_{13}>0\)
解析 由\(S_6=S_7\),则 \(\dfrac{6\left(a_1+a_6\right)}{2}=\dfrac{7\left(a_1+a_7\right)}{2}\),即\(a_1+6d=a_7=0\),
又\(S_5<S_6\), \(S_7>S_8\),易知:\(d<0\),
故数列\(\{a_n \}\)是递减数列,故\(C\)正确;
根据\(a_6+a_8=2a_7=0\),\(A\)正确;
\(S_5=\dfrac{5\left(a_1+a_5\right)}{2}=5 a_3\), \(S_8=\dfrac{8\left(a_1+a_8\right)}{2}=\dfrac{8\left(a_2+a_7\right)}{2}=4 a_2\),
则\(5a_3-4a_2=a_1+6d=0\),
故\(S_5=S_8\),故\(B\)正确;
再根据 \(S_{13}=\dfrac{13\left(a_1+a_{13}\right)}{2}=13 a_7=0\),故\(D\)错误,
故选:\(D\).
【巩固练习】
1.若\(\{a_n \}\)是等差数列,且\(a_2\),\(a_{2022}\)是方程\(x^2-4x+3=0\)的两个根,则 \(2^{a_1} \times 2^{a_2} \times 2^{a_3} \times \ldots \times 2^{a_{2023}}=\)( )
A. \(34\) \(\qquad \qquad \qquad \qquad\) B. \(35\) \(\qquad \qquad \qquad \qquad\) C. \(18\) \(\qquad \qquad \qquad \qquad\) D.\(19\)
2.已知等差数列\(\{a_n \}\)中,\(a_5\),\(a_{17}\)是方程\(x^2-6x-21=0\)的两根,则\(\{a_n \}\)的前\(21\)项的和为( )
A. \(6\) \(\qquad \qquad \qquad \qquad\) B. \(30\) \(\qquad \qquad \qquad \qquad\) C. \(63\) \(\qquad \qquad \qquad \qquad\) D.\(126\)
3.已知公差非零的等差数列\(\{a_n \}\)满足\(|a_5 |=|a_8 |\),则下列结论正确的是( )
A. \(S_{13}=0\)
B. 当\(S_{13}>0\)时,\(S_n≥S_6\),\(n∈N^*\)
C.当\(S_{13}<0\)时, \(S_n≥S_6\),\(n∈N^*\)
D. \(S_n=S_{13-n}\left(1 \leq n \leq 12, n \in N^*\right)\)
4.已知数列\(\{a_n \}\)、\(\{b_n \}\)都是等差数列,设\(\{a_n \}\)的前\(n\)项和为\(S_n\),\(\{b_n \}\)的前\(n\)项和为\(T_n\).若 \(\dfrac{S_n}{T_n}=\dfrac{2 n+1}{3 n+2}\),则 \(\dfrac{a_5}{b_5}=\) ( )
A. \(\dfrac{19}{29}\) \(\qquad \qquad \qquad \qquad\) B. \(\dfrac{11}{25}\) \(\qquad \qquad \qquad \qquad\) C. \(\dfrac{11}{17}\) \(\qquad \qquad \qquad \qquad\) D. \(\dfrac{2}{3}\)
参考答案
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答案 \(C\)
解析 \(\because a_2\),\(a_{2022}\)是方程\(x^2-4x+3=0\)的两个根,
\(\therefore a_2+a_{2022}=4\),
又\(\{a_n \}\)是等差数列,
\(\therefore a_1+a_{2023}=a_2+a_{2022}=\cdots=2 a_{1012}\),即 \(a_{1012}=2\),
\(2^{a_1} \times 2^{a_2} \times 2^{a_3} \times \ldots \times 2^{a_{2023}}=2^{\left(a_1+a_2+\cdots+a_{2023}\right)}=2^{(1011 \times 4+2)}=2^{4046}\).
故选:\(C\). -
答案 \(C\)
解析 \(\because\) 等差数列\(\{a_n \}\)中,\(a_5\),\(a_{17}\)是方程\(x^2-6x-21=0\)的两根,
\(\therefore a_5+a_{17}=6\), \(a_5 \cdot a_{17}=-21\),
故\(\{a_n \}\)的前\(21\)项的和为 \(\dfrac{21 \times\left(a_1+a_{21}\right)}{2}=\dfrac{21 \times\left(a_5+a_{17}\right)}{2}=63\),
故选:\(C\). -
答案 \(B\)
解析 \(\because\)公差非零的等差数列\(\{a_n \}\)满足\(|a_5 |=|a_8 |\),
\(\therefore a_5=-a_8\),即\(a_5+a_8=0\),即\(a_7+a_6=0\).
\(\therefore S_{12}=\dfrac{12\left(a_1+a_{12}\right)}{2}=6\left(a_5+a_8\right)=0\),故\(A\)不对;
对于选项\(B\), \(S_{13}=\dfrac{13 \times\left(a_1+a_{13}\right)}{2}=13 a_7>0\)时, \(\therefore a_7>0\), \(\therefore a_6<0\),
故该数列为递增等差数列,前\(6\)项为负数,从第\(7\)项开始为正数,故\(S_6\)最小,
即\(S_n≥S_6\),故\(B\)正确;
对于选项\(C\), \(S_{13}=\dfrac{13 \times\left(a_1+a_{13}\right)}{2}=13 a_7<0\)时, \(\therefore a_7<0\),\(\therefore a_6>0\),
故该数列为递减等差数列,前\(6\)项为正数,从第\(7\)项开始为负数,故\(S_6\)最大,
即\(S_n≤S_6\),故\(C\)错误;
当\(n=1\)时,\(S_n=S_1=a_1\), \(S_{13-n}=S_{12}=\dfrac{12 \times\left(a_1+a_{12}\right)}{2}=6\left(a_7+a_6\right)=0\),
而\(a_1≠0\),故\(D\)错误,
故选:\(B\). -
答案 \(A\)
解析 \(\because S_n\),\(T_n\)分别为等差数列\(\{a_n \}\)、\(\{b_n \}\)的前\(n\)项和, \(\dfrac{S_n}{T_n}=\dfrac{2 n+1}{3 n+2}\),
\(\therefore \dfrac{a_5}{b_5}=\dfrac{\dfrac{9\left(a_1+a_9\right)}{2}}{\dfrac{9\left(b_1+b_9\right)}{2}}=\dfrac{S_9}{T_9}=\dfrac{2 \times 9+1}{3 \times 9+2}=\dfrac{19}{29}\),
故选:\(A\).
【题型3】等差数列综合
【典题1】 设数列\(\{a_n \}\)的前\(n\)项和为\(S_n\),且 \(\left(S_n-1\right)^2=a_n S_n\).
(1)求\(a_1\);\(\qquad \qquad\) (2)求证:数列 \(\left\{\dfrac{1}{S_n-1}\right\}\)为等差数列.
解析 (1)解:\(n=1\)时, \(\left(a_1-1\right)^2=a_1^2\),则 \(a_1=\dfrac{1}{2}\);
(2)证明: \(\left(S_n-1\right)^2=a_n S_n\),
则\(n≥2\)时, \(\left(S_n-1\right)^2=\left(S_n-S_{n-1}\right) S_n\).
即有\(-2S_n+1=-S_{n-1} S_n\),
即\(1-S_n=S_n (1-S_{n-1})\),即有 \(\dfrac{1}{S_{n-1}-1}=\dfrac{S_n}{S_n-1}\),
\(\dfrac{1}{S_{n-1}}-\dfrac{1}{S_{n-1}-1}=\dfrac{1}{S_n-1}-\dfrac{S_n}{S_n-1}=-1\)为定值,
则数列 \(\left\{\dfrac{1}{S_n-1}\right\}\)为等差数列.
【巩固练习】
1.已知\(\{a_n \}\)是等差数列,\(a_4+a_6=8\),其前\(5\)项和\(S_5=40\).
(1)求\(\{a_n \}\)的通项\(a_n\);
(2)求\(\{a_n \}\)前\(n\)项和\(S_n\)的最大值.
2.流行性感冒(简称流感)是由流感病毒引起的急性呼吸道传染病.某市去年\(11\)月份曾发生流感,据资料统计,\(11\)月\(1\)日,该市新的流感病毒感染者有\(20\)人,此后,每天的新感染者平均比前一天的新感染者增加\(50\)人.由于该市医疗部门采取措施,使该种病毒的传播得到控制,从某天起,每天的新感染者平均比前一天的新感染者减少\(30\)人,到\(11\)月\(30\)日止,该市在这\(30\)天内感染该病毒的患者总共有\(8 670\)人,则\(11\)月几日,该市感染此病毒的新患者人数最多?并求这一天的新患者人数.
3.设\(S_n\)是数列\(\{a_n \}\)的前n项和且\(n∈N^*\),所有项\(a_n>0\),且 \(S_n=\dfrac{1}{4} a_n^2+\dfrac{1}{2} a_n-\dfrac{3}{4}\).
(1)证明:\(\{a_n \}\)是等差数列;\(\qquad \qquad\) (2)求数列\(\{a_n \}\)的通项公式.
4.已知正项数列\(\{a_n \}\)的首项\(a_1=1\),前\(n\)项和\(S_n\)满足 \(2 a_n=\sqrt{S_n}+\sqrt{S_{n-1}}(n \geq 2)\).
(1)求数列\(\{a_n \}\)的通项公式;
(2)记数列 \(\left\{\dfrac{1}{a_n a_{n+1}}\right\}\)的前\(n\)项和为\(T_n\),若对任意的\(n∈N^*\),不等式\(5T_n<a^2-a\)恒成立,求实数\(a\)的取值范围.
参考答案
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答案 (1) \(a_n=-2n+14\) ;(2) \(42\) .
解析 (1)由题意可得 \(\left\{\begin{array}{l} a_4+a_6=2 a_1+8 d=8 \\ S_5=5 a_1+10 d=40 \end{array}\right.\),解得\(a_1=12\),\(d=-2\),
\(\therefore a_n=12-2(n-1)=-2n+14\);
(2) \(S_n=\dfrac{n(12+14-2 n)}{2}=n(13-n)=-n^2+13 n=-\left(n-\dfrac{13}{2}\right)^2+\dfrac{169}{4}\),
当\(n=6\)或\(n=7\)时,有最大值,最大值为\(-36+13×6=42\). -
答案 \(11\)月\(12\)日该市感染此病毒的新患者人数最多,且这一天患者人数为\(570\).
解析 设\(11\)月\(1\)日,该市第\(n\)日\((n∈N^*,1≤n≤30)\)感染此病毒的新患者人数最多.
则从\(11\)月\(1\)日至第\(n\)日,每日感染此病毒的新患者人数构成一个等差数列,其首项为\(20\),公差为\(50\).
前\(n\)日患者总人数\(S_n=20 n+\dfrac{n(n-1)}{2} \times 50=25 n^2-5 n\).
从第\(n+1\)日开始至\(11\)月\(30\)日止,每日感染此病毒的新患者人数依次构成另一个等差数列.
其首项为\(20+(n-1)×50-30=50n-60\),公差为\(-30\).项数为\((30-n)\),
其患者总人数为 \(T_{30-n}=(30-n)(50 n-60)+\dfrac{(30-n)(29-n)}{2} \times(-30)\).
由题意可得 \(S_n+T_{30-n}=8670\),即\(\left(25 n^2-5 n\right)+\left(-65 n^2+2445 n-14850\right)=8670\).
化为\(n^2-61n+588=0\),解得\(n=12(1≤n≤30)\).
\(\therefore n=12\),第\(12\)日的新患者人数为\(20+(12-1)×50=570\).
\(\therefore\) \(11\)月\(12\)日该市感染此病毒的新患者人数最多,且这一天患者人数为\(570\). -
答案 (1) 略;(2) \(a_n=2n+1\).
解析 (1)因为 \(S_n=\dfrac{1}{4} a_n^2+\dfrac{1}{2} a_n-\dfrac{3}{4}\).
所以\(4S_n=a_n^2+2a_n-3\),\(4S_{n+1}=a_{n+1}^2+2a_{n+1}-3\),
两式相减整理可得\((a_{n+1}+a_n)( a_{n+1}-a_n-2)=0\),
\(\because a_n>0\),
\(\therefore a_{n+1}-a_n-2=0\),
\(\therefore a_{n+1}-a_n=2\),
\(\{a_n \}\)成等差数列;
(2)由(1)可知数列\(\{a_n \}\)是等差数列,并且\(4S_1=a_1^2+2a_1-3\),
所以\(a_1=3\)或\(-1\)(舍去),公差为\(2\),
所以\(a_n=2n+1\). -
答案 (1) \(a_n=\left\{\begin{array}{l} \dfrac{2 n+1}{4}, n \geq 2 \\ 1, n=1 \end{array}\right.\);(2)\(a≤-3\)或\(a≥4\).
解析 (1)当\(n≥2\)时, \(2 a_n=\sqrt{S_n}+\sqrt{S_{n-1}}\),
\(\therefore 2\left(S_n-S_{n-1}\right)=\sqrt{S_n}+\sqrt{S_{n-1}}\),即 \(\sqrt{S_n}-\sqrt{S_{n-1}}=\dfrac{1}{2}\),
所以数列 \(\left\{\sqrt{S_n}\right\}\)是首项为\(1\),公差为 \(\dfrac{1}{2}\)的等差数列,
故\(\sqrt{S_n}=\dfrac{n+1}{2}\), \(a_n=\dfrac{1}{2}\left(\sqrt{S_n}+\sqrt{S_{n-1}}\right)=\dfrac{1}{2} \cdot\left(\dfrac{n+1}{2}+\dfrac{n}{2}\right)=\dfrac{2 n+1}{4}(n \geq 2)\),
因此\(a_n=\left\{\begin{array}{l} \dfrac{2 n+1}{4}, n \geq 2 \\ 1, n=1 \end{array}\right.\).
(2)当\(n≥2\)时, \(\dfrac{1}{a_n a_{n+1}}=\dfrac{1}{\dfrac{2 n+1}{4} \cdot \dfrac{2 n+3}{4}}=8\left(\dfrac{1}{2 n+1}-\dfrac{1}{2 n+3}\right)\),
\(\therefore T_n=\dfrac{4}{5}+8\left(\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\cdots+\dfrac{1}{2 n+1}-\dfrac{1}{2 n+3}\right)=\dfrac{12}{5}-\dfrac{8}{2 n+3}<\dfrac{12}{5}\),
又\(\because 5T_n<a^2-a\),\(\therefore 12≤a^2-a\),解得\(a≤-3\)或\(a≥4\).
即所求实数\(a\)的范围是\(a≤-3\)或\(a≥4\).
分层练习
【A组---基础题】
1.已知数列\(\{a_n \}\)满足\(a_{n+1}-a_n=2\),若\(a_2=4\),则\(a_8=\) ( )
A.\(14\) \(\qquad \qquad \qquad \qquad\) B.\(16\) \(\qquad \qquad \qquad \qquad\) C.\(18\) \(\qquad \qquad \qquad \qquad\) D.\(20\)
2.若数列\(a\),\(x_1\),\(x_2\),\(b\)与\(a\),\(y_1\),\(y_2\),\(y_3\),\(b\)均为等差数列(其中\(a≠b\)),则 \(\dfrac{x_2-x_1}{y_2-y_1}=\)( )
A. \(\dfrac{2}{3}\) \(\qquad \qquad \qquad \qquad\) B. \(\dfrac{4}{3}\) \(\qquad \qquad \qquad \qquad\) C. \(\dfrac{3}{2}\) \(\qquad \qquad \qquad \qquad\) D. \(\dfrac{3}{4}\)
3.记\(S_n\)为等差数列\(\{a_n \}\)的前\(n\)项和,已知\(S_4=0\),\(a_5=5\),则( )
A. \(na_n<2S_n\) \(\qquad \qquad \qquad\) B. \(na_n=2S_n\) \(\qquad \qquad \qquad\) C.\(na_n>2S_n\) \(\qquad \qquad \qquad\) D.\(2na_n=S_n\)
4.我国古代数学著作《九章算术》中有如下问题:“今有善走男,日增等里,首日行走一百里,九日共行一千二百六十里,问日增几何?”,该问题中,善走男第\(5\)日所走的路程里数是( ).
A. \(110\) \(\qquad \qquad \qquad \qquad\) B. \(120\) \(\qquad \qquad \qquad \qquad\) C. \(130\) \(\qquad \qquad \qquad \qquad\) D.\(140\)
5.已知正项等差数列\(\{a_n \}\)的前\(n\)项和\(S_n\),\(S_9=45\),则\(a_2⋅a_8\)的最大值是 .
A. \(40\) \(\qquad \qquad \qquad \qquad\) B. \(50\) \(\qquad \qquad \qquad \qquad\) C.\(80\) \(\qquad \qquad \qquad \qquad\) D.\(25\)
6.在等差数列\(\{a_n \}\)中,前\(n\)项和\(S_n\),且 \(\dfrac{S_8}{S_4}=3\),则 \(\dfrac{S_{12}}{S_8}=\)\(\underline{\quad \quad}\).
7.设\(S_n\)是等差数列\(\{a_n \}\)的前\(n\)项和,存在\(\therefore n∈N^*\)且\(n>4\)时有\(S_8=20\), \(S_{2 n-1}-S_{2 n-9}=116\),则\(a_n=\)\(\underline{\quad \quad}\).
8.设等差数列\(\{a_n \}\)满足:\(a_1=3\),公差\(d∈(0,10)\),其前\(n\)项和为\(S_n\).若数列 \(\left\{\sqrt{S_n+1}\right\}\)也是等差数列,则 \(\dfrac{S_n+10}{a_n+1}\)的最小值为\(\underline{\quad \quad}\).
9.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\(a_n\).
10.数列\(\{a_n \}\)的首项\(a_1=-20\),\(a_n+a_{n+1}=3n-54\),\(n∈N^*\),
(1)求数列\(\{a_n \}\)的通项公式;
(2)设\(\{a_n \}\)的前\(n\)项和为\(S_n\),求\(S_n\)的最小值.
11.已知数列\(\{a_n \}\)的前n项和 \(S_n=4-a_n-\dfrac{1}{2^{n-2}}\) .
(1)证明:数列\(\{2^n a_n\}\)是等差数列;\(\qquad \qquad\) (2)求\(\{a_n \}\)的通项公式.
参考答案
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答案 \(B\)
解析 \(\because\)数列\(\{a_n \}\)满足\(a_{n+1}-a_n=2\), \(\therefore\)数列\(\{a_n \}\)是等差数列,公差为\(2\),
\(\because a_2=4\),\(\therefore a_8=a_2+(8-2)×2=4+12=16\),
故选:\(B\). -
答案 \(B\)
解析 设数列\(a\),\(x_1\),\(x_2\),\(b\)的公差为\(d_1\),数列\(a\),\(y_1\),\(y_2\),\(y_3\),\(b\)的公差为\(d_2\),
则有\(3d_1=b-a\),即 \(d_1=\dfrac{b-a}{3}\),且\(4d_2=b-a\),即 \(d_2=\dfrac{b-a}{4}\),
\(\therefore \dfrac{x_2-x_1}{y_2-y_1}=\dfrac{d_1}{d_2}=\dfrac{\dfrac{b-a}{3}}{\dfrac{b-a}{4}}=\dfrac{4}{3}\),故选:\(B\). -
答案 \(C\)
解析 设等差数列\(\{a_n \}\)的首项为\(a_1\),公差为\(d\),
由\(S_4=0\),\(a_5=5\),得 \(\left\{\begin{array}{l} 4 a_1+\dfrac{4 \times 3}{2} d=0 \\ a_1+4 d=5 \end{array}\right.\),解得\(a_1=-3\),\(d=2\).
\(\therefore a_n=-3+2 (n-1)=2n-5\), \(S_n=-3 n+\dfrac{n(n-1)}{2} \times 2=n^2-4 n\),
则\(na_n-2S_n=2n^2-5n-2n^2+8n=3n>0\),\(\therefore na_n>2S_n\),
\(2na_n=4n^2-10n≠S_n\).
故选:\(C\). -
答案 \(D\)
解析 \(\because\)今有善走男,日增等里,首日行走一百里,九日共行一千二百六十里,
设善走男每天走的路程为\(\{a_n \}\),则数列\(\{a_n \}\)为等差数列,设公差为\(d\),则\(a_1=100\),
\(\therefore\)由题意, \(9 a_1+\dfrac{9 \times 8}{2} \cdot d=1260\),可得\(a_1+4d=140\),
解得该善走男第\(5\)日所走的路程里数为\(a_5=a_1+4d=140\),
故选:\(D\). -
答案 \(D\)
解析 正项等差数列\(\{a_n \}\)的前\(n\)项和\(S_n\),\(S_9=45\),
则 \(\dfrac{9\left(a_1+a_9\right)}{2}=45\),化为\(a_1+a_9=10=a_2+a_8\).
则 \(a_2 \cdot a_8 \leq\left(\dfrac{a_2+a_8}{2}\right)^2=25\),当且仅当\(a_2=a_8=5\)时取等号. -
答案 \(2\)
解析 等差数列\(\{a_n \}\)中,前\(n\)项和\(S_n\),且\(\dfrac{S_8}{S_4}=3\),
设\(S_4=t\),则\(S_8=3t\),
\(\therefore S_4\),\(S_8-S_4\),\(S_{12}-S_8\)也是等差数列,
即\(t\),\(2t\),\(S_{12}-S_8\)成等差数列,
故有 \(S_{12}-S_8=3t\),故有\(S_{12}=S_8+3t=6t\),
则 \(\dfrac{S_{12}}{S_8}=\dfrac{6 t}{3 t}=2\). -
答案 \(\dfrac{17}{2}\)
解析 由题知\(a_1+a_2+⋯+a_8=20\),
且 \(S_{2 n-1}-S_{2 n-9}=a_{2 n-8}+a_{2 n-7}+\cdots+a_{2 n-1}=116\),
故\(a_1+a_{2 n-1}=\dfrac{20+116}{8}=17=2 a_n\),
所以 \(a_n=\dfrac{17}{2}\). -
答案 \(3\)
解析 由题意可得: \(2 \sqrt{S_2+1}=\sqrt{S_1+1}+\sqrt{S_3+1}\),
即\(2 \sqrt{7+d}=2+\sqrt{10+3 d}\),公差\(d∈(0,10)\),解得\(d=2\).
\(\therefore a_n=2n+1\).
\(\therefore S_n=\dfrac{n(3+2 n+1)}{2}=n^2+2 n\).
\(\therefore \sqrt{S_n+1}=n+1\).
\(\therefore\)数列 \(\left\{\sqrt{S_n+1}\right\}\)是等差数列,
则 \(\dfrac{S_n+10}{a_n+1}=\dfrac{n^2+2 n+10}{2 n+2}=\dfrac{(n+1)^2+9}{2(n+1)}=\dfrac{1}{2}\left[(n+1)+\dfrac{9}{n+1}\right]\)\(\geq \sqrt{(n+1) \cdot \dfrac{9}{n+1}}=3\),
当且仅当\(n=2\)时取等号,
\(\therefore \dfrac{S_n+10}{a_n+1}\)的最小值为\(3\). -
答案 证明略, \(a_n=\dfrac{2 n-1}{2 n}\)
解析 证明:\(\because a_{n+1}-a_n=2(a_{n+1}-1)( a_n-1)\),
\(\therefore 2(a_{n+1}-1)( a_n-1)=( a_{n+1}-1)-(a_n-1)\),
上式两边同除以\((a_{n+1}-1)( a_n-1)\) \((\)可验证\((a_{n+1}-1)( a_n-1)≠0)\),
化简得\(\dfrac{1}{a_{n+1}-1}-\dfrac{1}{a_n-1}=-2\),
所以\(\left\{\dfrac{1}{a_n-1}\right\}\)是以\(-2\)为首项,\(-2\)为公差的等差数列,
即\(\dfrac{1}{a_n-1}=-2-2(n-1)=-2 n\),
即\(a_n=1-\dfrac{1}{2 n}=\dfrac{2 n-1}{2 n}\). -
答案 (1) \(a_n=\left\{\begin{array}{l} \dfrac{3}{2} n-\dfrac{43}{2}, n \text { 为奇数 } \\ \dfrac{3}{2} n-34, n \text { 为偶数 } \end{array}\right.\); (2)\(-243\).
解析 (1)\(a_1=-20\),\(a_2=-31\),
又\(a_{n+1}+a_{n+2}=3n-51\),\(a_n+a_{n+1}=3n-54\),
则\(a_{n+2}-a_n=3\),即奇数项成等差,偶数项成等差,且公差均为\(3\),
\(\therefore a_n=\left\{\begin{array}{l} \dfrac{3}{2} n-\dfrac{43}{2}, n \text { 为奇数 } \\ \dfrac{3}{2} n-34, n \text { 为偶数 } \end{array}\right.\);
(2)当\(n\)为偶数,即\(n=2k\)时\(S_n=-51 k+\dfrac{k(k-1)}{2} \times 6=3(k-9)^2-243\),
\(\therefore S_n≥S_{18}=-243\);
当\(n\)为奇数,即\(n=2k-1\)时\(S_n=S_{2 k}-a_{2 k}=3\left(k-\dfrac{19}{2}\right)^2-236 \dfrac{3}{4}\),
\(\therefore S_n \geq S_{17}=S_{19}=-236\),
\(\therefore\left(S_n\right)_{\min }=S_{18}=-243\). -
答案 (1)略;(2) \(a_n=\dfrac{n}{2^{n-1}}\) .
解析 证明 (1)\(\because\)数列\(\{a_n \}\)的前\(n\)项和 \(S_n=4-a_n-\dfrac{1}{2^{n-2}}\) .
\(\therefore\)当\(n=1\)时, \(S_1=4-a_1-\dfrac{1}{2^{-1}}\),解得\(a_1=1\),
当\(n≥2\)时, \(S_n=4-a_n-\dfrac{1}{2^{n-2}}\), \(S_{n-1}=4-a_{n-1}-\dfrac{1}{2^{n-3}}\) .
两式相减,得 \(2 a_n=a_{n-1}+\dfrac{4}{2^n}\),
\(\therefore 2 \times 2^n a_n=2 \times 2^{n-1} a_{n-1}+4\),
\(\therefore 2^n a_n-2^{n-1} a_{n-1}=\dfrac{2 \times 2^{n-1} a_{n-1}+4}{2}-2^{n-1} a_{n-1}\)\(=2^{n-1} a_{n-1}+2-2^{n-1} a_{n-1}=2\),
又\(2a_1=2\),
\(\therefore\)数列\(\{2^n a_n\}\)是首项为\(2\),公差为\(2\)的等差数列.
(2)\(\because\)数列\(\{2^n a_n\}\)是首项为\(2\),公差为\(2\)的等差数列,
\(\therefore 2^n a_n=2+(n-1)×2=2n\),
\(\therefore a_n=\dfrac{2 n}{2^n}=\dfrac{n}{2^{n-1}}\).
\(\therefore \{a_n\}\)的通项公式为 \(a_n=\dfrac{n}{2^{n-1}}\).
【B组---提高题】
1.已知等差数列\(\{a_n \}\),\(S_n\)是数列\(\{a_n \}\)的前\(n\)项和,对任意的\(n∈N^*\),均有\(S_4≥S_n\)成立,则 \(\dfrac{a_{10}}{a_6}\)的值不可能是( )
A. \(2\) \(\qquad \qquad \qquad \qquad\) B. \(3\) \(\qquad \qquad \qquad \qquad\) C.\(4\) \(\qquad \qquad \qquad \qquad\) D.\(5\)
2.设\(S_n\)是数列\(\{a_n \}\)的前\(n\)项和,已知\(a_1=1\),\(a_n=-S_n S_{n-1} (n≥2)\),则\(S_n=\)\(\underline{\quad \quad}\).
3.记\(S_n\)为数列\(\{a_n \}\)的前\(n\)项和,\(b_n\)为数列\(\{S_n\}\)的前\(n\)项积,已知 \(\dfrac{2}{S_n}+\dfrac{1}{b_n}=2\).
(1)证明:数列\(\{b_n\}\)是等差数列;\(\qquad \qquad\) (2)求\(\{a_n \}\)的通项公式.
参考答案
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答案 \(A\)
解析 根据题意,等差数列\(\{a_n \}\),对任意的\(n∈N^*\),均有\(S_4≥S_n\)成立,
即\(S_4\)是等差数列\(\{a_n \}\)的前\(n\)项和中的最大值,
必有\(a_1>0\),公差\(d<0\),
分\(3\)种情况讨论:
①\(a_4=0\)时,\(S_3=S_4\),\(S_3\),\(S_4\)是等差数列\(\{a_n \}\)的前n项和中的最大值,
\(a_4=a_1+3d=0\),则\(a_1=-3d\),
\(\dfrac{a_{10}}{a_6}=\dfrac{a_1+9 d}{a_1+5 d}=\dfrac{6 d}{2 d}=3\);
②\(a_5=0\)时, \(S_4=S_5\), \(S_4\),\(S_5\)是等差数列\(\{a_n \}\)的前\(n\)项和中的最大值,
\(a_5=a_1+4d=0\),则\(a_1=-4d\),
\(\dfrac{a_{10}}{a_6}=\dfrac{a_1+9 d}{a_1+5 d}=\dfrac{5 d}{d}=5\);
③\(a_4>0\),\(a_5<0\)时,\(S_4\)是等差数列\(\{a_n \}\)的前\(n\)项和中的最大值,
此时\(a_4=a_1+3d>0\),\(a_5=a_1+4d<0\),
则\(-3d<a_1<-4d\),变形可得 \(-4<\dfrac{a_1}{d}<-3\),
\(\dfrac{a_{10}}{a_6}=\dfrac{a_1+9 d}{a_1+5 d}=\dfrac{\dfrac{a_1}{d}+9}{\dfrac{a_1}{d}+5}=1+\dfrac{4}{\dfrac{a_1}{d}+5}\),
\(\because-4<\dfrac{a_1}{d}<-3\), \(\therefore 3<\dfrac{a_{10}}{a_6}<5\),
综上, \(3 \leq \dfrac{a_{10}}{a_6} \leq 5\).
故选:\(A\). -
答案 \(S_n=\dfrac{2}{n+1}\)
解析 \(\because 2a_n=-S_n S_{n-1} (n≥2)\), \(\therefore 2(S_n-S_{n-1})=-S_n S_{n-1}\),
\(\therefore \dfrac{1}{S_n}-\dfrac{1}{S_{n-1}}=\dfrac{1}{2}\),
\(\because a_1=1\), \(\therefore \dfrac{1}{S_1}=1\),
\(\therefore\left\{\dfrac{1}{S_n}\right\}\)是以\(1\)为首项, \(\dfrac{1}{2}\)为公差的等差数列
\(\therefore \dfrac{1}{S_n}=1+\dfrac{1}{2}(n-1)=\dfrac{n+1}{2}\),
\(\therefore S_n=\dfrac{2}{n+1}\). -
答案 (1)略 ;(2) \(a_n= \begin{cases}\dfrac{3}{2}, & n=1 \\ -\dfrac{1}{n(n+1)}, & n \geq 2\end{cases}\).
解析 (1)证明:当\(n=1\)时,\(b_1=S_1\),
由 \(\dfrac{2}{b_1}+\dfrac{1}{b_1}=2\),解得 \(b_1=\dfrac{3}{2}\),
当\(n≥2\)时, \(\dfrac{b_n}{b_{n-1}}=S_n\),代入 \(\dfrac{2}{S_n}+\dfrac{1}{b_n}=2\),
消去\(S_n\),可得 \(\dfrac{2 b_{n-1}}{b_n}+\dfrac{1}{b_n}=2\),所以 \(b_n-b_{n-1}=\dfrac{1}{2}\),
所以\(\{b_n\}\)是以\(\dfrac{3}{2}\)为首项,\(\dfrac{1}{2}\)为公差的等差数列.
(2)由题意,得\(a_1=S_1=b_1=\dfrac{3}{2}\),
由(1),可得 \(b_n=\dfrac{3}{2}+(n-1) \times \dfrac{1}{2}=\dfrac{n+2}{2}\),
由 \(\dfrac{2}{S_n}+\dfrac{1}{b_n}=2\),可得 \(S_n=\dfrac{n+2}{n+1}\),
当\(n≥2\)时, \(a_n=S_n-S_{n-1}=\dfrac{n+2}{n+1}-\dfrac{n+1}{n}=-\dfrac{1}{n(n+1)}\),显然\(a_1\)不满足该式,
所以\(a_n= \begin{cases}\dfrac{3}{2}, & n=1 \\ -\dfrac{1}{n(n+1)}, & n \geq 2\end{cases}\).
【C组---拓展题】
1.设\(a\),\(b\),\(c\)分别是\(△ABC\)内角\(A\),\(B\),\(C\)的对边,若 \(\dfrac{1}{\tan A}\), \(\dfrac{1}{\tan B}\), \(\dfrac{1}{\tan C}\)依次成公差不为\(0\)的等差数列,则( )
A.\(a\),\(b\),\(c\)依次成等差数列 \(\qquad \qquad \qquad \qquad \qquad\) B.\(a^2\),\(b^2\),\(c^2\)依次成等差数列
C.\(\sqrt{a}\), \(\sqrt{b}\), \(\sqrt{c}\)依次成等差数列 \(\qquad \qquad \qquad \quad\) D.\(\dfrac{1}{a}\), \(\dfrac{1}{b}\),\(\dfrac{1}{c}\)依次成等差数列
2.已知数列\(\{a_n \}\)满足\(a_1=6\),\(a_2=-3\),\(a_n+a_{n+3}=a_{n+1}+a_{n+2}\),\(n∈N^*\).
(1)若\(a_3=4\),求\(a_4\),\(a_5\)的值;
(2)证明:对任意正实数\(m\),\(\{a_{2n}+ma_{2n+1}\}\)成等差数列;
(3)若\(a_n>a_{n+1} (n∈N^*)\),\(a_3+a_4=-33\),求数列\(\{a_n \}\)的通项公式.
参考答案
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答案 \(B\)
解析 \(\because \dfrac{1}{\tan A}\), \(\dfrac{1}{\tan B}\), \(\dfrac{1}{\tan C}\)依次成公差不为\(0\)的等差数列,
\(\therefore \dfrac{2}{\tan B}=\dfrac{1}{\tan A}+\dfrac{1}{\tan C}\), \(\therefore \dfrac{2 \cos B}{\sin B}=\dfrac{\cos A}{\sin A}+\dfrac{\cos C}{\sin C}\),
\(\because \dfrac{\cos A}{\sin A}+\dfrac{\cos C}{\sin C}=\dfrac{\sin C \cos A+\cos C i \sin A}{\sin A \sin C}=\dfrac{\sin (A+C)}{\sin A \sin C}=\dfrac{\sin B}{\sin A \sin C}\),
\(\therefore \dfrac{2 \cos B}{\sin B}=\dfrac{\sin B}{\sin A \sin n C}\),
\(\therefore 2 \cos B=\dfrac{\sin ^2 B}{\sin A \sin C}=\dfrac{b^2}{a c}\),即 \(2 a c \cos B=b^2\),
\(\therefore a^2+c^2-b^2=b^2\), \(\therefore a^2+c^2=2b^2\),
\(\therefore a^2\),\(b^2\),\(c^2\)成等差数列,
因此只有\(B\)正确.
故选:\(B\). -
答案 (1) \(a_4=-5\),\(a_5=2\);(2)略 ;(3) \(a_n=15-9n,n∈N^*\).
解析 (1)解:数列\(\{a_n \}\)满足\(a_1=6\),\(a_2=-3\),\(a_n+a_(n+3)=a_{n+1}+a_{n+2}\).
当\(n=1\)时,\(a_1+a_4=a_2+a_3\), \(\therefore a_4=-5\),
当\(n=2\)时,\(a_2+a_5=a_3+a_4\),\(\therefore a_5=2\).
(2)证明:\(\because\) 数列\(\{a_n \}\)满足\(a_1=6\),\(a_2=-3\),\(a_n+a_{n+3}=a_{n+1}+a_{n+2},n∈N^*\).①
\(\therefore n≥2\)时,\(a_{n-1}+a_{n+2}=a_n+a_{n+1}\),②
①+②,得\(a_{n-1}+a_{n+3}=2a_{n+1}\), \(\therefore a_{n+3}-a_{n+1}=a_{n+1}-a_{n-1}\),
\(\therefore\left\{a_{2 n-1}\right\}\)是等差数列,设公差为\(d_1\),\(\{a_{2n}\}\)是等差数列,设公差为\(d_2\),
\(\therefore\left(a_{2 n}+m a_{2 n+3}\right)-\left(a_{2 n}+m a_{2 n+1}\right)=\left(a_{2 n+2}-a_{2 n}\right)+m\left(a_{2 n+3}-a_{2 n+1}\right)=d_2+m d_1\),
\(\therefore\)对任意正实数\(m\), \(\left\{a_{2 n}+m a_{2 n+1}\right\}\)成等差数列.
(3)解:设奇数项所成等差数列的公差为\(d_1\),偶数项所成等左数列的公差为\(d_2\),
①当\(n\)为奇数时, \(a_n=6+\dfrac{n-1}{2} d_1\), \(a_{n+1}=-3+\dfrac{n-1}{2} d_2\),
则 \(6+\dfrac{n-1}{2} d_1>-3+\dfrac{n-1}{2} d_2\),即\(n(d_1-d_2)+18+2(d_2-d_1)>0\),
\(\therefore\left\{\begin{array}{l} d_1-d_2 \geq 0 \\ 1 \times\left(d_1-d_2\right)+9+d_2-d_1>0 \end{array}\right.\),\(\therefore d_1-d_2≥0\).
②当\(n\)为偶数时, \(a_n=-3+\left(\dfrac{n}{2}-1\right) d_2\), \(a_{n+1}=6+\dfrac{n}{2} d_1\),
则 \(-3+\left(\dfrac{n}{2}-1\right) d_2>6+\dfrac{n}{2} d_1\), \(\therefore n(d_1-d_2)+18+2d_2<0\),
\(\therefore\left\{\begin{array}{l} d_1-d_2 \leq 0 \\ 2 \times\left(d_1-d_2\right)+18+2 d_2<0 \end{array}\right.\), \(\therefore\left\{\begin{array}{l} d_1-d_2 \leq 0 \\ d_1<-9 \end{array}\right.\),
综上,得\(d_1=d_2<-9\),
\(\because a_3+a_4=a_1+a_2+d_1+d_2=3+2d_1=-33\),解得\(d_1=-18\),
\(\therefore\) 当\(n\)为奇数时, \(a_n=6+\dfrac{n-1}{2} \times(-18)=15-9 n\),
当\(n\)为偶数时, \(a_n=-3+\left(\dfrac{n}{2}-1\right) \times(-18)=15-9 n\),
\(\therefore\)数列\(\{a_n \}\)的通项公式为\(a_n=15-9n,n∈N^*.\)