4.1 数列的概念2 (递推公式、前n项和)
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选择性第二册同步巩固,难度2颗星!
基础知识
递推公式
若已知数列\(\{a_n\}\)的第一项\(a_1\)(或前\(n\)项),且任一项\(a_n\)和它的前一项\(a_{n-1}\)(或前\(n\)项)间的关系可以用一公式表示,那么这个公式叫做这个数列的递推公式.
解释
(1) 举例:\(a_1\)(初始条件),\(a_n=2a_{n-1}+3n(n≥2)\)(递推关系);
\(a_1=1\) ,\(a_2=2\)(初始条件) ,\(a_n=3a_{n-1}-2a_{n-2} (n≥3)\)(递推关系).
(2) 通项公式与递推公式的异同
不同点 | 相同点 | |
通项公式 | 可根据某项的序号,直接用代入法求出该项 | 都可确定一个数列,都可求出数列的任何一项 |
递推公式 | 可根据第1项或前几项的值,通过一次或多次赋值逐项求出数列的项,直至求出所需的项 |
【例1】你能写出满足数列\(1,1,2,3,5,8…\)的一个递推公式么?
答案 \(a_1=1\) ,\(a_2=1\) ,\(a_n=a_{n-1}+a_{n-2} (n≥3)\).
【例2】已知数列\(\{a_n\}\)中,\(a_1=1\),\(a_{n+1}=3a_n-1\),则\(a_3\)等于 .
解 \(a_2=3a_1-1=2\),\(a_3=3a_2-1=5.\)
an与Sn的关系
若\(S_n\)为数列\(\{a_n\}\)的前n项和,即\(S_n=a_1+a_2+...+a_n\).
则 \(a_n= \begin{cases}S_1 & , n=1 \\ S_n-S_{n-1} & , n \geq 2\end{cases}\).
解释
(1) 若已知列\(\{a_n\}\)的前\(n\)项和\(S_n\),可利用公式\(a_n= \begin{cases}S_1 & , n=1 \\ S_n-S_{n-1} & , n \geq 2\end{cases}\)求数列通项公式\(a_n\),
(2) 证明 若\(S_n\)为数列\(\{a_n\}\)的前\(n\)项和,根据定义可得,\(S_1=a_1\),
\(S_n=a_1+a_2+...+a_{n-1} +a_n\) (1),\(S_{n-1}=a_1+a_2+...+a_{n-1}\) (2).
故当\(n=1\)时,\(a_n=a_1=S_1\);
当\(n≥2\)时,由(1)-(2)得\(S_n-S_{n-1}=a_n\),
即 \(a_n= \begin{cases}S_1 & , n=1 \\ S_n-S_{n-1} & , n \geq 2\end{cases}\).
基本方法
【题型1】 递推公式
【典题1】 已知数列\(\{a_n\}\)满足\(a_1=1\), \(a_n=a_{n-1}+\dfrac{1}{n(n-1)}(n \geq 2)\),写出该数列前\(5\)项,并归纳出它的一个通项公式.
解析 \(a_1=1\), \(a_2=a_1+\dfrac{1}{2 \times 1}=1+\dfrac{1}{2}=\dfrac{3}{2}\), \(a_3=a_2+\dfrac{1}{3 \times 2}=\dfrac{3}{2}+\dfrac{1}{6}=\dfrac{5}{3}\),
\(a_4=a_3+\dfrac{1}{4 \times 3}=\dfrac{5}{3}+\dfrac{1}{12}=\dfrac{7}{4}\), \(a_5=a_4+\dfrac{1}{5 \times 4}=\dfrac{7}{4}+\dfrac{1}{20}=\dfrac{9}{5}\).
故数列的前\(5\)项分别为 \(1, \dfrac{3}{2}, \dfrac{5}{3}, \dfrac{7}{4}, \dfrac{9}{5}\).
由于 \(1=\dfrac{2 \times 1-1}{1}\), \(\dfrac{3}{2}=\dfrac{2 \times 2-1}{2}\), \(\dfrac{5}{3}=\dfrac{2 \times 3-1}{3}\),\(\dfrac{7}{4}=\dfrac{2 \times 4-1}{4}\),\(\dfrac{9}{5}=\dfrac{2 \times 5-1}{5}\),
故数列\(\{a_n\}\)的一个通项公式为 \(a_n=\dfrac{2 n-1}{n}=2-\dfrac{1}{n}\).
【巩固练习】
1.下列给出的图形中,星星的个数构成一个数列,则该数列的一个递推公式可以是( )
A.\(a_{n+1}=a_n+n,n∈N^*\) \(\qquad \qquad \qquad \qquad \qquad\) B.\(a_n=a_{n-1}+n,n∈N^*,n⩾2\)
C.\(a_{n+1}=a_n+(n+1),n∈N^*,n⩾2\) \(\qquad \qquad \qquad\) D.\(a_n=a_{n-1}+(n-1),n∈N^*,n⩾2\)
2.数列\(\{a_n\}\)中,\(a_2=1\),且\(a_{n+1}=na_n\),则\(a_3=\)\(\underline{\quad \quad}\).
3.在数列\(\{a_n\}\)中,已知\(a_1=2\),\(a_2=7\),\(a_{n+2}\)等于\(a_n a_{n+1} (n∈N^*)\)的个位数,则\(a_{2017}=\)\(\underline{\quad \quad}\).
4.在数列\(\{x_n \}\)中, \(\dfrac{2}{x_n}=\dfrac{1}{x_{n-1}}+\dfrac{1}{x_{n+1}}(n \geqslant 2)\),且 \(x_2=\dfrac{2}{3}\), \(x_4=\dfrac{2}{5}\),则\(x_{10}=\)\(\underline{\quad \quad}\).
5.设数列\(\{a_n\}\)满足\(a_{n+1}=a_n^2-na_n+1\),\(n=1,2,3,…\),\(a_1=2\),通过求\(a_2\),\(a_3\)猜想\(a_n\)的一个通项公式为\(\underline{\quad \quad}\).
6.已知数列\(\{a_n\}\)的第一项是\(1\),以后各项由公式\(a_{n-1}=2a_n-2(n>1)\)给出,写出这个数列的前\(5\)项,并猜下它的一个通项公式(不需要证明).
参考答案
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答案 \(B\)
解析 根据题意,可得\(a_1=1\),\(a_2=3\),\(a_3=6\),\(a_4=10\),
发现规律: \(a_n=\dfrac{n(n+1)}{2}\),
而 \(a_{n+1}-a_n=\dfrac{(n+1)(n+2)}{2}-\dfrac{n(n+1)}{2}=\dfrac{n+1}{2}[(n+2)-n]=n+1\)
故\(a_{n+1}=a_n+n+1\)成立,
即\(a_n=a_{n-1}+n,n∈N^*,n⩾2\),
故选:\(B\). -
答案 \(2\)
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答案 \(2\)
解析 数列\(\{a_n\}\),\(a_1=2\),\(a_2=7\),\(a_{n+2}\)等于\(a_n a_{n+1} (n∈N^*)\)的个位数,
\(\therefore a_2\cdot a_1=14\),\(\therefore a_3=4\),
同理可得:\(a_4=8,a_5=2,a_6=6,a_7=2,a_8=2,a_9=4,a_{10}=8,…\),
所以数列从第\(3\)项开始呈现周期性出现,周期为\(6\),即\(a_{n+6}=a_n,n≥3\),
则\(a_{2017}=a_{335×6+7}=a_7=2\). -
答案 \(\dfrac{2}{11}\)
解析 由于在数列\(\{x_n \}\)中, \(\dfrac{2}{x_n}=\dfrac{1}{x_{n-1}}+\dfrac{1}{x_{n+1}}(n \geqslant 2)\),且 \(x_2=\dfrac{2}{3}\), \(x_4=\dfrac{2}{5}\),
则 \(\dfrac{2}{x_3}=\dfrac{1}{x_2}+\dfrac{1}{x_4}=\dfrac{3}{2}+\dfrac{5}{2}=4\),故 \(x_3=\dfrac{2}{4}\),
同理得到 \(x_5=\dfrac{2}{6}\),所以 \(x_n=\dfrac{2}{n+1}\),故得到 \(x_{10}=\dfrac{2}{11}\),
故答案为 \(\dfrac{2}{11}\). -
答案 \(a_n=n+1\)
解析 \(\because a_1=2\),\(a_{n+1}=a_n^2-na_n+1\),
\(\therefore a_2=a_1^2-a_1+1=3\),\(a_3=a_2^2-2a_2+1=4\),\(a_4=a_3^2-3a_3+1=5\),
故猜想\(a_n=n+1\). -
答案 \(a_1=1\), \(a_2=\dfrac{3}{2}\), \(a_3=\dfrac{7}{4}\), \(a_4=\dfrac{15}{8}\), \(a_5=\dfrac{31}{16}\), \(a_n=\dfrac{2^n-1}{2^{n-1}}\) .
解析 \(\because a_{n-1}=2a_n-2(n>1)\),
\(\therefore a_n=1+\dfrac{1}{2} a_{n-1}(n>1)\).
又\(a_1=1\), \(\therefore a_2=1+\dfrac{1}{2} a_1=1+\dfrac{1}{2} \times 1=\dfrac{3}{2}\),
\(a_3=1+\dfrac{1}{2} a_2=1+\dfrac{1}{2} \times \dfrac{3}{2}=\dfrac{7}{4}\), \(a_4=1+\dfrac{1}{2} a_3=1+\dfrac{1}{2} \times \dfrac{7}{4}=\dfrac{15}{8}\),
\(a_5=1+\dfrac{1}{2} a_5=1+\dfrac{1}{2} \times \dfrac{15}{8}=\dfrac{31}{16}\),
\(\therefore\) 这个数列的前\(5\)项是\(a_1=1\), \(a_2=\dfrac{3}{2}\), \(a_3=\dfrac{7}{4}\), \(a_4=\dfrac{15}{8}\), \(a_5=\dfrac{31}{16}\),
可归纳得数列的通项公式\(a_n=\dfrac{2^n-1}{2^{n-1}}\) .
【题型2】 通项公式与前n项和
【典题1】 已知下面各数列\(\{a_n\}\)的前\(n\)项和\(S_n\)的公式,求\(\{a_n\}\)}的通项公式.
(1) \(S_n=2n^2-3n\); \(\qquad \qquad\) (2) \(S_n=3^n-2\).
解析 (1)当\(n=1\)时,\(a_1=S_1=2×1^2-3×1=-1\);
当\(n≥2\)时,\(S_{n-1}=2(n-1)^2-3(n-1)=2n^2-7n+5\),
则\(a_n=S_n-S_{n-1}=(2n^2-3n)-(2n^2-7n+5)=2n^2-3n-2n^2+7n-5=4n-5\).
此时若\(n=1\),则\(a_n=4n-5=4×1-5=-1=a_1\),
故\(a_n=4n-5\).
(2)当\(n=1\)时,\(a_1=S_1=3^1-2=1\);
当\(n≥2\)时,\(S_{n-1}=3^(n-1)-2\),
则\(a_n=S_n-S_{n-1}=(3^n-2)-(3^{n-1}-2)=3^n-3^{n-1}=3⋅3^{n-1}-3^{n-1}=2⋅3^{n-1}\).
\(a_1=1\)不满足\(a_n=2⋅3^{n-1}\),
故 \(a_n= \begin{cases}1, & n=1 \\ 2 \cdot 3^{n-1}, & n \geq 2\end{cases}\).
点拨 若题目中已知数列前\(n\)项和\(S_n\),可利用公式\(a_n= \begin{cases}S_1 & , n=1 \\ S_n-S_{n-1} & , n \geq 2\end{cases}\)求数列的通项公式\(a_n\),注意分类讨论,最后要检验\(a_1\)是否满足\(a_n=f(n)(n≥2)\).
【典题2】 已知数列\(\{a_n\}\)的前\(n\)项和为\(S_n=n^2⋅a_n (n⩾2)\),而\(a_1=1\),通过计算\(a_2,a_3,a_4\),猜想\(a_n\)等于( )
A. \(\dfrac{2}{(n+1)^2}\) \(\qquad \qquad \qquad \qquad\) B. \(\dfrac{2}{n(n+1)}\) \(\qquad \qquad \qquad \qquad\) C. \(\dfrac{1}{2^n-1}\) \(\qquad \qquad \qquad \qquad\) D. \(\dfrac{1}{2 n-1}\)
解析 (1)\(\because S_n=n^2 a_n\),\(\therefore a_{n+1}=S_(n+1)-S_n=(n+1)^2 a_{n+1}-n^2 a_n\)
\(\therefore a_{n+1}=\dfrac{n}{n+2} a_n\),
\(\therefore a_2=\dfrac{1}{1+2}=\dfrac{1}{3}\), \(a_3=\dfrac{2}{2+2} \cdot \dfrac{1}{3}=\dfrac{1}{6}\),
猜测 \(a_n=\dfrac{2}{n(n+1)}\),
故选:\(B\).
【巩固练习】
1.已知数列\(\{a_n\}\)的前n项和为\(S_n\),若\(S_n =\dfrac{1}{n}\),\(n∈N^*\),则\(a_2=\)( )
A. \(-\dfrac{1}{2}\) \(\qquad \qquad \qquad \qquad\) B. \(-\dfrac{1}{6}\) \(\qquad \qquad \qquad \qquad\) C. \(\dfrac{1}{6}\) \(\qquad \qquad \qquad \qquad\) D. \(\dfrac{1}{2}\)
2.在一个数列中,如果每一项与它的后一项的和为同一个常数,我们把这个数列叫做“等和数列”,这个常数叫做该数列的公和,已知数列\(\{a_n\}\)是等和数列且\(a_1=1\),公和为\(4\),则数列\(\{a_n\}\)的前\(n\)项和\(S_n\)的计算公式为\(S_n=\)\(\underline{\quad \quad}\).
3.已知数列\(\{a_n\}\)的通项公式为\(a_n=\dfrac{1}{(n+1) \sqrt{n}+n \sqrt{n+1}}\left(n \in N^*\right)\),其前\(n\)项和为\(S_n\),则在数列\(S_1,S_2,…,S_{2019}\)中,有理数项的项数为\(\underline{\quad \quad}\).
4.已知数列\(\{a_n\}\)的前\(n\)项和\(S_n=n^2+2\),求此数列的通项公式.
5.已知数列\(\{a_n\}\)的前n项和\(S_n\)满足\(S_n=\dfrac{1}{3}\left(a_n-1\right)\left(n \in N^*\right)\).
(1)求\(a_1,a_2,a_3,a_4\);
(2)由\(a_1,a_2,a_3,a_4\)的值猜想这个数列的通项公式(不用证明).
参考答案
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答案 \(A\)
解析 根据题意,数列\(\{a_n\}\)的前\(n\)项和 \(S_n =\dfrac{1}{n}\),
则\(a_2=S_2-S_1=\dfrac{1}{2}-1=-\dfrac{1}{2}\);
故选:\(A\). -
答案 \(S_n=\left\{\begin{array}{l} 2 n, n \text { 为偶数 } \\ 2 n-1, n \text { 为奇数 } \end{array}\right.\)
解析 由题意知,\(a_n+a_{n+1}=4\),且\(a_1=1\),
所以\(a_1+a_2=4\),得\(a_2=3,a_3=1,a_4=3\),
则\(a_n=\left\{\begin{array}{l} 1, n \text { 为奇数 } \\ 3, n \text { 为偶数 } \end{array}\right.\),
当\(n\)为偶数时\(S_n=(1+3)+(1+3)+(1+3)+⋯+(1+3)=4×\dfrac{n}{2}=2n\),
当\(n\)为奇数时\(S_n=(1+3)+(1+3)+⋯(1+3)+1=4×\dfrac{n-1}{2}+1=2n-1\),
故\(S_n=\left\{\begin{array}{l} 2 n, n \text { 为偶数 } \\ 2 n-1, n \text { 为奇数 } \end{array}\right.\). -
答案 \(43\)
解析 由题意,可知: \(a_n=\dfrac{1}{(n+1) \sqrt{n}+n \sqrt{n+1}}=\dfrac{(n+1) \sqrt{n}-n \sqrt{n+1}}{[(n+1) \sqrt{n}+n \sqrt{n+1}][(n+1) \sqrt{n}-n \sqrt{n+1}]}\)
\(=\dfrac{(n+1) \sqrt{n}-n \sqrt{n+1}}{n(n+1)}=\dfrac{\sqrt{n}}{n}-\dfrac{\sqrt{n+1}}{n+1}\).
\(\therefore S_n=a_1+a_2+\cdots+a_n=1-\dfrac{\sqrt{2}}{2}+\dfrac{\sqrt{2}}{2}-\dfrac{\sqrt{3}}{3}+\cdots+\dfrac{\sqrt{n}}{n}-\dfrac{\sqrt{n+1}}{n+1}=1-\dfrac{\sqrt{n+1}}{n+1}\).
\(\therefore S_3, S_8, S_{15} \cdots\)为有理项,
又\(\because\) 下标\(3,8,15,…\)的通项公式为\(b_n=n^2-1(n⩾2)\),
\(\therefore n^2-1⩽2019\),且\(n⩾2\),解得:\(2⩽n⩽44\),
\(\therefore\) 有理项的项数为\(44-1=43\). -
答案 \(a_n=\left\{\begin{array}{l} 3, n=1 \\ 2 n-1, n \geq 2 \end{array}\right.\)
解析 当\(n≥2\)时,\(a_n=S_n-S_{n-1}=n^2+2-(n-1)^2-2=2n-1\);
当\(n=1\)时,\(a_1=S_1=1^2+2=3\),不适合上式,
故 \(a_n=\left\{\begin{array}{l} 3, n=1 \\ 2 n-1, n \geq 2 \end{array}\right.\). -
答案 (1) \(a_1=-\dfrac{1}{2}, \quad a_2=\dfrac{1}{4}, \quad a_3=-\dfrac{1}{8}, \quad a_4=\dfrac{1}{16}\);(2) \(a_n=\left(-\dfrac{1}{2}\right)^n\).
解析 (1) \(\because S_1=a_1\), \(S_n=\dfrac{1}{3}\left(a_n-1\right)\),
\(\therefore a_1=\dfrac{1}{3}\left(a_1-1\right)\), \(\therefore a_1=-\dfrac{1}{2}\),
\(\therefore S_2=\dfrac{1}{3}\left(a_2-1\right)=a_1+a_2\), \(\therefore a_2=\dfrac{1}{4}\),
同理可得 \(a_3=-\dfrac{1}{8}, \quad a_4=\dfrac{1}{16}\),
(2)由\(a_1,a_2,a_3,a_4\)的值猜想这个数列的通项公式 \(a_n=\left(-\dfrac{1}{2}\right)^n\).
分层练习
【A组---基础题】
1.在数列\(\{a_n\}\)中, \(a_1=\dfrac{1}{2}\), \(a_{n+1}=1-\dfrac{1}{a_n}\) ,则\(a_5=\)( )
A.\(2\) \(\qquad \qquad \qquad \qquad\) B.\(3\) \(\qquad \qquad \qquad \qquad\) C.\(-1\) \(\qquad \qquad \qquad \qquad\) D. \(\dfrac{1}{2}\)
2.设数列\(\{a_n\}\)满足\(a_1=2\), \(a_{n+1}=1-\dfrac{1}{a_n}\left(n \in \boldsymbol{N}^*\right)\),则\(a_{2019}=\)( )
A.\(2\) \(\qquad \qquad \qquad \qquad\) B.\(\dfrac{1}{2}\) \(\qquad \qquad \qquad \qquad\) C.\(-\dfrac{1}{2}\) \(\qquad \qquad \qquad \qquad\) D.\(-1\)
3.已知数列\(\{a_n\}\)满足:\(a_1=m\),\(m\)为正整数, \(a_{n+1}=\left\{\begin{array}{l}
\dfrac{a_n}{2}, \text { 当 } a_n \text { 为偶数时 } \\
3 a_n+1, \text { 当 } a_n \text { 为奇数时 }
\end{array}\right.\),若\(a_6=1\),则\(m\)所有可能的取值为( )
A.\(\{4,5\}\) \(\qquad \qquad \qquad \qquad\) B.\(\{4,32\}\) \(\qquad \qquad \qquad \qquad\) C.\(\{4,5,32\}\) \(\qquad \qquad \qquad \qquad\) D.\(\{5,32\}\)
4.已知数列\(\{a_n\}\)的前\(n\)项和\(S_n=3^n (λ-n)-6\),若数列\(\{a_n\}\)单调递减,则\(λ\)的取值范围是( )
A.\((-∞,2)\) \(\qquad \qquad \qquad \qquad\) B.\((-∞,3)\) \(\qquad \qquad \qquad \qquad\) C.\((-∞,4)\) \(\qquad \qquad \qquad \qquad\) D.\((-∞,5)\)
5.(多选)定义“等积数列”:在一个数列中,如果每一项与它的后一项的积都为同一个常数,那么这个数列叫做等积数列,这个常数叫做该数列的公积,已知数列\(\{a_n\}\)是等积数列,且\(a_1=3\),前\(7\)项的和为\(14\),则下列结论正确的是( )
A.\(a_{n+2}=a_n\) \(\qquad \qquad \qquad\) B. \(a_2=\dfrac{2}{3}\) \(\qquad \qquad \qquad\) C.公积为\(1\) \(\qquad \qquad \qquad\) D.\(a_n a_{n+1} a_{n+2}=6\)
6.已知数列\(\{a_n\}\)中,\(a_1=3\),\(a_{n+1}=3a_n+2\),则\(a_3=\)\(\underline{\quad \quad}\).
7.已知数列\(\{a_n\}\)满足:\(a_1=2\), \(a_{n+1}=\dfrac{a_n-1}{a_n}\),猜想数列\(\{a_n\}\)的前\(2014\)项的和\(S_{2014}=\)\(\underline{\quad \quad}\).
8.若数列\(\{a_n\}\)的前\(n\)项和\(S_n=2n^2-3n+2\),则它的通项公式\(a_n\)是\(\underline{\quad \quad}\).
9.已知数列\(\{a_n\}\)的通项公式和为 \(S_n=\dfrac{n(7 n+3)}{2}\),\(n∈N^*\),现从前\(m\)项:\(a_1,a_2,…,a_m\)中抽出一项(不是 也不是\(a_m\),余下各项的算术平均数为\(40\),则抽出的是第\(\underline{\quad \quad}\)项.
10.数列\(\{a_n\}\)的前n项和是\(S_n\).若\(2S_n=na_n+2(n⩾2,n∈N^* )\),\(a_2=2\),则\(a_1=\)\(\underline{\quad \quad}\);\(a_n=\)\(\underline{\quad \quad}\).
11.数列\(\{a_n\}\)前\(n\)项和为\(S_n\),且\(S_n=an^2+bn+c(a,b,c∈R)\),已知\(a_1=-28\),\(S_2=-52\),\(S_5=-100\).
(1)求数列\(\{a_n\}\)的通项公式.\(\qquad \qquad\) (2)求使得\(S_n\)最小的序号\(n\)的值.
12.(1)在数列\(\{a_n\}\)中,\(a_1=1\), \(a_{n+1}=\dfrac{2 a_n}{2+a_n}\),\(n∈N^*\).猜想这个数列的通项公式.
(2)已知正项数列\(\{a_n\}\)的前\(n\)项和\(S_n\),满足 \(S_n=\dfrac{1}{2}\left(a_n+\dfrac{1}{a_n}\right)\)\((n∈N^* )\),求出\(a_1,a_2,a_3\),并推测\(a_n\)的表达式.
参考答案
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答案 \(C\)
解析 \(a_1=\dfrac{1}{2}\), \(a_{n+1}=1-\dfrac{1}{a_n}\) ,
则\(a_2=1-2=-1\),\(a_3=1+1=2\),\(a_4=1-\dfrac{1}{2}=\dfrac{1}{2}\),\(a_5=1-2=-1\),
故选:\(C\). -
答案 \(D\)
解析 数列\(\{a_n\}\)满足\(a_1=2\), \(a_{n+1}=1-\dfrac{1}{a_n}\left(n \in \boldsymbol{N}^*\right)\),
则\(a_2=1-\dfrac{1}{2}=\dfrac{1}{2}\),同理可得:\(a_3=-1\),\(a_4=2\),\(a_5=\dfrac{1}{2}\),\(…\)
可得\(a_{n+3}=a_n\).
\(a_{2019}=a_3 \times 672+3=a_3=-1\).
故选:\(D\). -
答案 \(C\)
解析 \(\because a_6=1\),
\(\therefore a_5\)必为偶数, \(\therefore a_6=\dfrac{a_5}{2}=1\),解得\(a_5=2\).
当\(a_4\)为偶数时, \(a_5=\dfrac{a_4}{2}\),解得\(a_4=4\);
当\(a_4\)为奇数时,\(a_5=3a_4+1=2\),解得\(a_4=\dfrac{1}{3}\),舍去.
\(\therefore a_4=4\).
当\(a_3\)为偶数时, \(a_4=\dfrac{a_3}{2}=4\),解得\(a_3=8\);
当\(a_3\)为奇数时,\(a_4=3a_3+1=4\),解得\(a_3=1\).
当\(a_3=8\)时,当\(a_2\)为偶数时, \(a_3=\dfrac{a_2}{2}=8\),解得\(a_2=16\);
当\(a_2\)为奇数时,\(a_3=3a_2+1=8\),解得 \(a_2=\dfrac{7}{3}\),舍去.
当\(a_3=1\)时,当\(a_2\)为偶数时, \(a_3=\dfrac{a_2}{2}=1\),解得\(a_2=2\);
当\(a_2\)为奇数时,\(a_3=3a_2+1=1\),解得\(a_2=0\),舍去.
当\(a_2=16\)时,当\(a_1\)为偶数时,\(a_2=\dfrac{a_1}{2}=16\),解得\(a_1=32=m\);
当\(a_1\)为奇数时,\(a_2=3a_1+1=16\),解得\(a_1=5=m\).
当\(a_2=2\)时,当\(a_1\)为偶数时,\(a_2=\dfrac{a_1}{2}=2\),解得\(a_1=4=m\);
当\(a_1\)为奇数时,\(a_2=3a_1+1=2\),解得\(a_1=\dfrac{1}{3}\),舍去.
综上可得\(m=4,5,32\).
故选:\(C\). -
答案 \(A\)
解析 \(\because S_n=3^n (λ-n)-6\),① \(\therefore S_{n-1}=3^{n-1} (λ-n+1)-6,n>1\),②
①-②得数列\(a_n=3^{n-1} (2λ-2n-1)(n>1,n∈N^*)\)为单调递减数列,
\(\therefore a_n>a_{n+1}\),且\(a_1>a_2\),
\(\therefore 3^{n-1} (2λ-2n-1)>3^n (2λ-2n-3)\),且\(λ<2\),
化为\(λ<n+2,(n>1)\),且\(λ<2\),
\(\therefore λ<2\),
\(\therefore λ\)的取值范围是\((-∞,2)\).
故选:\(A\). -
答案 \(AB\)
解析 设\(a_n a_{n+1}=k\)(\(k\)为常数),则\(a_{n+1} a_{n+2}=k\),
所以 \(\dfrac{a_{n+2}}{a_n}=1\),即\(a_{n+2}=a_n\),故\(A\)正确,
\(\because\)前\(7\)项的和为\(14\),\(\therefore 3(a_1+a_2 )+a_1=14\),
\(\because a_1=3\), \(\therefore a_2=\dfrac{2}{3}\),
\(\therefore a_n a_{n+1}=2\),即公积为\(2\),故\(B\)正确,\(C\)错误,
当\(n\)为奇数时,\(a_n a_{n+1} a_{n+2}=6\);
当\(n\)为偶数时, \(a_n a_{n+1} a_{n+2}=\dfrac{4}{3}\),故\(D\)错误.
故选:\(AB\). -
答案 \(35\)
解析 \(a_1=3\),则\(a_2=3a_1+2=3×3+2=11\),\(a_3=3a_2+2=3×11+2=35\). -
答案 \(\dfrac{2017}{2}\)
解析 \(\because a_1=2\),\(a_{n+1}=\dfrac{a_n-1}{a_n}\),
\(\therefore a_2=\dfrac{2-1}{2}=\dfrac{1}{2}\), \(a_3=\dfrac{\dfrac{1}{2}-1}{\dfrac{1}{2}}=-1\), \(a_4=\dfrac{-1-1}{-1}=2\),
\(\therefore\)数列\(\{a_n\}\)是以\(3\)为周期的数列,
又 \(S_3=a_1+a_2+a_3=2+\dfrac{1}{2}-1=\dfrac{3}{2}\),
\(\therefore S_{2014}=S_{2013}+a_{2014}=671 \times \dfrac{3}{2}+a_1=\dfrac{2013}{2}+2=\dfrac{2017}{2}\).
故答案为: \(\dfrac{2017}{2}\). -
答案 \(a_n=\left\{\begin{array}{l} 1, n=1 \\ 4 n-5, n \geq 2 \end{array}\right.\)
解析 当\(n=1\)时,\(a_1=S_1=2-3+2=1\).
当\(n≥2\)时,\(a_n=S_n-S_{n-1}=2n^2-3n+2-[2(n-1)^2-3(n-1)+2]=4n-5\).
\(\therefore a_n=\left\{\begin{array}{l} 1, n=1 \\ 4 n-5, n \geq 2 \end{array}\right.\). -
答案 \(6\)
解析 设抽出的一项是第\(x\)项,由题得, \(a_n= \begin{cases}S_1 & n=1 \\ S_n-S_{n-1}=7 n-2 & n \geqslant 2\end{cases}\),
且\(S_m=40(m-1)+a_x\),
\(\therefore \dfrac{m(7 m+3)}{2}=40(m-1)+7 x-2\),
\(\therefore m^2-11m+12-2x=0\),
\(\therefore x=6\)时,\(m=11,m=0\)(舍去),
\(\therefore\) 抽出的是第\(6\)项.
故答案为:\(6\). -
答案 \(a_1=1\), \(a_n=\left\{\begin{array}{l} 1, n=1 \\ 2 n-2, n \geqslant 2 \end{array}\right.\).
解析 当\(n=2\)时,\(\because 2(a_1+a_2 )=2a_2+2\),\(\therefore a_1=1\),
\(\therefore\)当\(n⩾2\)时,有\(2S_{n-1}=(n-1)a_{n-1}+2\),
\(\therefore 2a_n=na_n-(n-1)a_{n-1}\),即\((n-2)a_n=(n-1)a_{n-1}\),
\(\therefore\) 当\(n⩾3\)时,有 \(\dfrac{a_n}{a_{n-1}}=\dfrac{n-1}{n-2}\),
\(\therefore \dfrac{a_3}{a_2}=\dfrac{2}{1}\), \(\dfrac{a_4}{a_3}=\dfrac{3}{2}\), \(\dfrac{a_5}{a_4}=\dfrac{4}{3}\), \(\ldots\), \(\dfrac{a_n}{a_{n-1}}=\dfrac{n-1}{n-2}\),
以上\(n-2\)个式相乘得,
\(\dfrac{a_n}{a_2}=n-1\),\(\therefore a_n=2n-2\),
当\(n=2\)时\(a_2=2\)符合上式,
\(\therefore a_n=\left\{\begin{array}{l} 1, n=1 \\ 2 n-2, n \geqslant 2 \end{array}\right.\). -
答案 (1) \(a_n=4n-32\); (2)\(7\)或\(8\).
解析 (1)有题意可得 \(\left\{\begin{array}{l} a+b+c=-28 \\ 4 a+2 b+c=-52 \\ 25 a+5 b+c=100 \end{array}\right.\)解得 \(\left\{\begin{array}{l} a=2 \\ b=-30 \\ c=0 \end{array}\right.\),
\(\therefore S_n=2n^2-30n\),
因为当\(n⩾2\)时,\(a_n=S_n-S_{n-1}=4n-32\)
当\(n=1\)时,\(a_1=-28\),也适合上式.
\(\therefore a_n=4n-32\);
(2)因为 \(S_n=2 n^2-30 n=2\left(n-\dfrac{15}{2}\right)^2-\dfrac{225}{2}\)
因为\(n\)是正整数,所以当\(n=7\)或\(8\),\(S_n\)最小,最小值是\(-112\). -
答案 (1) \(a_n=\dfrac{2}{n+1}\); (2) \(a_n=\sqrt{n}-\sqrt{n-1}\).
解析 (1)\(a_1=1\), \(a_2=\dfrac{2 a_1}{2+a_1}=\dfrac{2}{3}\), \(a_3=\dfrac{2 a_2}{2+a_2}=\dfrac{1}{2}\), \(a_4=\dfrac{2 a_3}{2+a_3}=\dfrac{2}{5}\),
猜想 \(a_n=\dfrac{2}{n+1}\).
(2)\(n=1\)时, \(a_1=\dfrac{1}{2}\left(a_1+\dfrac{1}{a_1}\right)\),解得\(a_1=1\).
\(n=2\)时, \(1+a_2=\dfrac{1}{2}\left(a_2+\dfrac{1}{a_2}\right)\),解得 \(a_2=\sqrt{2}-1\).
\(n=3\)时, \(1+\sqrt{2}-1+a_3=\dfrac{1}{2}\left(a_3+\dfrac{1}{a_3}\right)\),解得 \(a_3=\sqrt{3}-\sqrt{2}\).
猜想 \(a_n=\sqrt{n}-\sqrt{n-1}\).
【B组---提高题】
1.设 \(0<\theta<\dfrac{\pi}{2}\),已知\(a_1=2 \cos \theta\), \(a_{n+1}=\sqrt{2+a_n}\left(n \in \boldsymbol{N}^*\right)\),猜想\(a_n=\)( )
A. \(2 \cos \dfrac{\theta}{2^n}\) \(\qquad \qquad \qquad\) B. \(2 \cos \dfrac{\theta}{2^{n-1}}\) \(\qquad \qquad \qquad\) C. \(2 \cos \dfrac{\theta}{2^{n+1}}\) \(\qquad \qquad \qquad\) D. \(2 \sin \dfrac{\theta}{2^n}\)
2.已知数列\(\{a_n\}\)中,\(a_1=t\), \(a_{n+1}=\dfrac{a_n}{2}+\dfrac{2}{a_n}\),若\(\{a_n\}\)为单调递减数列,则实数\(t\)的取值范围是\(\underline{\quad \quad}\) .
3.已知正项数列\(\{a_n\}\)满足\(a_1=1\),\(2a_n a_{n+1}+3a_{n+1}=8a_n-2\).试比较\(a_n\)与\(2\)的大小,并说明理由.
参考答案
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答案 \(B\)
解析 当\(n=1\)时,\(A\)选项 \(2 \cos \dfrac{\theta}{2^n}=2 \cos \dfrac{\theta}{2}\),\(\therefore\)排除\(A\).
当\(n=2\)时,\(C\)选项 \(2 \cos \dfrac{\theta}{2^{n+1}}=2 \cos \dfrac{\theta}{4}\),\(\therefore\)排除\(C\).
\(a_2=\sqrt{2+a_1}=\sqrt{2+2 \cos \theta}=\sqrt{4 \cos ^2 \dfrac{\theta}{2}}=2 \cos \dfrac{\theta}{2}\),
此时\(D\)选项 \(2 \sin \dfrac{\theta}{2^n}=2 \sin \dfrac{\theta}{4}\),
\(\therefore\) 排除\(D\).
故选:\(B\). -
答案 \((2,+∞)\)
解析 \(\because a_{n+1}=\dfrac{a_n}{2}+\dfrac{2}{a_n}\),
\(\therefore a_{n+1}-a_n=\dfrac{2}{a_n}-\dfrac{a_n}{2}=\dfrac{\left(2-a_n\right)\left(2+a_n\right)}{2 a_n}<0\),
解得\(a_n>2\)或\(-2<a_n<0\),
(1)\(a_1=t∈(-2,0)\)时, \(a_2=\dfrac{a_1}{2}+\dfrac{2}{a_1}<-2\),
归纳可得:\(a_n<-2(n≥2)\).
\(\therefore a_2-a_1<0\),但是\(a_{n+1}-a_n>0(n≥2)\),不合题意,舍去.
(2)\(a_1=t>2\)时, \(a_2=\dfrac{a_1}{2}+\dfrac{2}{a_1}>2\),
归纳可得:\(a_n>2(n≥2)\).\(\therefore a_{n+1}-a_n<0\),符合题意. -
答案 \(a_n<2\)
解析 \(\because\)正项数列\(\{a_n\}\)满足\(a_1=1\),\(2a_n a_{n+1}+3a_{n+1}=8a_n-2\).
\(\therefore a_{n+1}=\dfrac{8 a_n-2}{2 a_n+3}\),
\(\therefore a_{n+1}-2=\dfrac{8 a_n-2}{2 a_n+3}-2=\dfrac{4 a_n-8}{2 a_n+3}=\dfrac{4\left(a_n-2\right)}{2 a_n+3}\),
\(\because\)正项数列\(\{a_n\}\)中,\(a_n>0\),\(\therefore 2a_n+3>0\),
\(\therefore a_{n+1}-2\)与\(a_n-2\)同号,\(\therefore a_n-2\)与\(a_1-2\)同号,
\(\because a_1-2=-1<0\),\(\therefore a_n-2<0\),
\(\therefore a_n<2\).
【C组---拓展题】
1.已知正项数列\(\{a_n\}\)满足\(a_1=1\), \(a_n=a_{n+1}+\dfrac{1}{\sqrt{n}} a_{n+1}^2\).则下列正确的是( )
A. \(\dfrac{1}{a_{n+1}}-\dfrac{1}{a_n}>\dfrac{1}{\sqrt{n}}\) \(\qquad \qquad \qquad \qquad \qquad\) B.数列\(\{a_{n+1}-a_n\}\)是递减数列
C.数列\(\{a_{n+1}+a_n\}\)是递增数列 \(\qquad \qquad \qquad \qquad\) D. \(a_{n+1}>\sqrt{n+1}-\sqrt{n}\)
2.已知数列\(\{a_n\}\)满足\(a_1=t\), \(a_{n+1}=1+\dfrac{1}{a_n}\),数列\(\{a_n\}\)可以是无穷数列,也可以是有穷数列,如取\(t=1\)时,可得无穷数列:\(1,2, \dfrac{3}{2}, \dfrac{5}{3}, \ldots\);取\(t=-\dfrac{1}{2}\)时,可得有穷数列:\(-\dfrac{1}{2}\),\(-1\),\(0\).
(1)若\(a_5=0\),求\(t\)的值;
(2)若\(1<a_n<2\)对任意\(n≥2\),\(n∈N^*\)恒成立,求实数\(t\)的取值范围;
(3)设数列\(\{b_n\}\)满足\(b_1=-1\), \(b_{n+1}=\dfrac{1}{b_n-1}\left(n \in \boldsymbol{N}^*\right)\),求证:\(t\)取数列\(\{b_n\}\)中的任何一个数,都可以得到一个有穷数列\(\{a_n\}\).
参考答案
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答案 \(D\)
解析 因为\(a_1=1\), \(a_n=a_{n+1}+\dfrac{1}{\sqrt{n}} a_{n+1}^2\),
所以 \(a_{n+1}-a_n=-\dfrac{a_{n+1}{ }^2}{\sqrt{n}}<0\),即\(a_{n+1}<a_n\),
① \(a_n=a_{n+1}+\dfrac{1}{\sqrt{n}} a_{n+1}^2<a_{n+1}+\dfrac{a_{n+1} a_n}{\sqrt{n}}\),
所以 \(\dfrac{1}{a_{n+1}}<\dfrac{1}{a_n}+\dfrac{1}{\sqrt{n}}\),即 \(\dfrac{1}{a_{n+1}}-\dfrac{1}{a_n}<\dfrac{1}{\sqrt{n}}\),\(A\)错误;
②易得 \(a_{n+1}-a_n=-\dfrac{a_{n+1}{ }^2}{\sqrt{n}}\), \(a_{n+2}-a_{n+1}=-\dfrac{a_{n+2}^2}{\sqrt{n+1}}\),
故 \(\dfrac{a_{n+2}-a_{n+1}}{a_{n+1}-a_n}=\dfrac{\sqrt{n} a_{n+2}{ }^2}{\sqrt{n+1} a_{n+1}^2}=\sqrt{\dfrac{n}{n+1}} \cdot\left(\dfrac{a_{n+2}}{a_{n+1}}\right)^2<1\),
则\(a_{n+2}-a_{n+1}>a_{n+1}-a_n\),数列\(\{a_{n+1}-a_n\}\)是递增数列,\(B\)错误;
③ \(a_{n+1}+a_n=2 a_{n+1}+\dfrac{a_{n+1}^2}{\sqrt{n}}\), \(a_{n+1}+a_{n+2}=2 a_{n+2}+\dfrac{a_{n+2}^2}{\sqrt{n+1}}\),
所以\(\dfrac{a_{n+1}+a_{n+2}}{a_n+a_{n+1}}=\dfrac{a_{n+2}}{a_{n+1}} \cdot \dfrac{2+\dfrac{a_{n+2}}{\sqrt{n+1}}}{2+\dfrac{a_{n+1}}{\sqrt{n}}}\),
因为\(\dfrac{\dfrac{a_{n+2}}{\sqrt{n+1}}}{\dfrac{a_{n+1}}{\sqrt{n}}}=\dfrac{\sqrt{n}}{\sqrt{n+1}} \cdot \dfrac{a_{n+2}}{a_{n+1}}<1\),
所以\(\dfrac{a_{n+1}+a_{n+2}}{a_n+a_{n+1}}=\dfrac{a_{n+2}}{a_{n+1}} \cdot \dfrac{2+\dfrac{a_{n+2}}{\sqrt{n+1}}}{2+\dfrac{a_{n+1}}{\sqrt{n}}}<1\),\(C\)错误;
④当\(n≥2\)时, \(\dfrac{1}{a_{n+1}}-\dfrac{1}{a_n}<\dfrac{1}{\sqrt{n}}<\dfrac{2}{\sqrt{n+1}+\sqrt{n-1}}=\sqrt{n+1}-\sqrt{n-1}\),
\(\dfrac{1}{a_{n+1}}=\dfrac{1}{a_{n+1}}-\dfrac{1}{a_n}+\dfrac{1}{a_n}-\dfrac{1}{a_{n-1}}+\cdots+\dfrac{1}{a_2}-\dfrac{1}{a_1}+\dfrac{1}{a_1}<\sqrt{n+1}+\sqrt{n}\),
则 \(a_{n+1}>\dfrac{1}{\sqrt{n}+\sqrt{n+1}}=\sqrt{n+1}-\sqrt{n}\),\(D\)正确.
故选:\(D\). -
答案 (1) \(-\dfrac{3}{5}\);(2) \(t>1\);(3)略.
解析 (1)数列\(\{a_n\}\)满足\(a_1=t\), \(a_{n+1}=1+\dfrac{1}{a_n}\),
所以 \(a_n=\dfrac{1}{a_{n+1}-1}\),
所以 \(a_4=\dfrac{1}{0-1}=-1\), \(a_3=\dfrac{1}{-1-1}=-\dfrac{1}{2}\), \(a_2=\dfrac{1}{-\dfrac{1}{2}-1}=-\dfrac{2}{3}\),
所以 \(t=\dfrac{1}{-\dfrac{2}{3}-1}=-\dfrac{3}{5}\).
(2)若\(1<a_n<2\)对任意\(n≥2\),\(n∈N^*\)恒成立,
所以 \(\dfrac{1}{2}<\dfrac{1}{a_n}<1\), \(\dfrac{3}{2}<a_{n+1}=1+\dfrac{1}{a_n}<2\),
即\(a_1=t\), \(a_2=\dfrac{t+1}{t}\),
所以 \(1<\dfrac{t+1}{t}<2\),解得\(t>1\).
证明:(3)由于 \(b_n=1+\dfrac{1}{b_{n+1}}\) ,
设\(a_1=1=b_k\),
则 \(a_2=1+\dfrac{1}{b_k}=b_{k-1}\), \(a_3=1+\dfrac{1}{b_{k-1}}=b_{k-2}\),\(…\),
\(a_k=b_1=-1\), \(a_{k+1}=1+\dfrac{1}{-1}=0\),
故数列\(\{a_n\}\)有\(k+1\)项,且为有穷数列.