求数列的通项公式

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求数列的通项公式

知识剖析

求数列的通项公式是高考常考的一专题,形式多样,解题方法很多,常见的有累加法、累乘法、待定系数法、迭代法、取倒数法等,课外延申的还有不动点法等,不管什么方法,一定要理解解题方法的本质,清楚每种方法的适用范围,避免出现“看得懂,模仿做还行,独立思考就含糊”的情况.

经典例题

【方法一】观察法

\({\color{Red}{适用范围:}}\)给出数列的前几项,猜测通项公式;
\({\color{Red}{方法:}}\)通过观察,得知数列各项之间数值的关系(比如数值之间的差或商成一定规律)或数值结构特点(比如数值的正负,分式,平方)从而求得通项公式.

【典题1】写出下列数列\(\{a_n\}\)的一个通项公式
\((1)-7,14,-21,28,…\)
\((2) \dfrac{1}{4}, \dfrac{3}{8}, \dfrac{5}{16}, \dfrac{7}{32} \ldots\)
\((3) 2,5,10,17,26,…\)
\((4) 32,332,3332,33332,….\)
\((5) 1,2,2,3,3,4,4,….\)
【解析】 \({\color{Red}{分解结构法:}}\)注意数值的结构,看其是否可视为两个或多个数列组合而成.
\((1)\)数列\(-7 ,14 ,-21 ,28\),…每项可分解成符号和项的绝对值相乘得到,****
求数列的通项公式
故\(a_{n}=(-1)^{n} 7 n\);
\((2)\)数列\(\dfrac{1}{4}, \dfrac{3}{8}, \dfrac{5}{16}, \dfrac{7}{32} \ldots\)…每项可分解成分子和分母相除得到,
求数列的通项公式
故\(a_n=(2n-1)2^{n+1}\);
\({\color{Red}{变形法:}}\)数列本身特点不明显,但通过加减乘除某个数之类方式变形成“规律感更强”的数列.
\((3)\)数列\(2 ,5 ,10 ,17 ,26\),…中若每项减去\(1\),则变成\(1\),\(4\),\(9\),\(16\),\(25\),…,
这些数都是完全平方数,易想到数列的通项是\(n^2\),
则原数列只需要在这基础上加回\(1\)便可,即\(a_n=n^2+1\).
\((4)\)数列\(2\),\(32\),\(332\),\(3332\),\(33332\),….中若每项加上\(1\),则变成\(3\),\(33\),\(333\),\(3333\),\(33333\),….,再每项乘以\(3\),变成\(9\),\(99\),\(999\),\(9999\),\(99999\),…
其中\(9=10-1\),\(99=10^2-1\),\(999=10^3-1\),\(9999=10^4-1\),\(99999=10^5-1\),
则其通项\(b_n=10^{n+1}-1\),
要求原数列的通项公式,
则“逆回去”,除以\(3\)再减\(1\)可得\(a_{n}=\dfrac{b_{n}}{3}-1=\dfrac{10^{n+1}-1}{3}-1=\dfrac{10^{n+1}-4}{3}\).
\({\color{Red}{分奇偶项}}\)
\((5)\)数列\(1\),\(2\),\(2\),\(3\),\(3\),\(4\),\(4\),… , 相邻每项之间没什么关系,若分奇偶性来看,就简单多了,
可得奇数项为\(1\),\(2\),\(3\),\(4\),… , 可得\(a_{n}=\dfrac{n+1}{2}\).偶数项为\(2\),\(3\),\(4\),… , 可得\(a_{n}=\dfrac{n+2}{2}\).
则该数列通项公式\(a_{n}= \begin{cases}\dfrac{n+1}{2}, & n \text { 为奇数 } \\ \dfrac{n+2}{2}, & n \text { 为偶数 }\end{cases}\).
【点拨】观察法主要是依靠“数感”,以上讲解的“分解结构法”“变形法”可有助于观察,它对后面讲到的利用数学归纳法求解通项公式有用.

巩固练习

1(★)数列\(1,-\dfrac{\sqrt{2}}{2}, \dfrac{1}{2},-\dfrac{\sqrt{2}}{4}, \dfrac{1}{4}\), …的一个通项公式为(  )
A.\(\left(-\dfrac{1}{2}\right)^{n-1}\)
B.\(\left(-\dfrac{\sqrt{2}}{2}\right)^{n}\)
C.\((-1)^{n}\left(\dfrac{\sqrt{2}}{2}\right)^{n-1}\)
D.\((-1)^{n+1}\left(\dfrac{\sqrt{2}}{2}\right)^{n-1}\)

2(★)下列可作为数列\(1\),\(2\),\(1\),\(2\),\(1\),\(2\),…的通项公式的是(  )
A.\(a_{n}=\dfrac{1+(-1)^{n-1}}{2}\)
B.\(a_{n}=\dfrac{3+(-1)^{n}}{2}\)
C.\(a_{n}=2-\sin \dfrac{n \pi}{2}\)
D.\(a_{n}=2-\cos [(n-1) \pi]\)

3(★★)写出以下各数列的一个通项公式.
\((1)1,-\dfrac{1}{2}, \dfrac{1}{4},-\dfrac{1}{8}, \ldots\)
\((2)10 ,9 ,8 ,7 ,6 ,…\)
\((3)0 ,3 ,8 ,15 ,24 ,…\)
\((4) \dfrac{1}{2}, \dfrac{5}{6}, \dfrac{11}{12}, \dfrac{19}{20}, \dfrac{31}{30}\)
\((5)4 ,44 ,444 ,4444 ,….\)

答案

1.\(D\)
2.\(B\)
3.\((1) a_{n}=(-1)^{n+1} \times \dfrac{1}{2^{n-1}}\)

\((2) a_n=11-n\)
\((3) a_n=n^2-1\)
\((4)a_{n}=1-\dfrac{1}{n(n+1)}\)
\((5) a_{n}=\dfrac{4}{9} \times\left(10^{n}-1\right)\).

【方法二】a_n与S_n的关系公式法

\({\color{Red}{适用范围:}}\)若得知\(S_n\)或\(a_n\)与\(S_n\)的关系式,求数列通项公式.
\({\color{Red}{方法:}}\)利用\(a_n\)与\(S_n\)的关系\(a_{n}=\left\{\begin{array}{cc} S_{1} & n=1 \\ S_{n}-S_{n-1} & n \geq 2 \end{array}\right.\),注意分类讨论,最后确定\(a_1\)是否满足\(a_n=f(n)\)\((n≥2)\).

【典题1】已知数列\(\{a_n\}\)的前\(n\)项和\(S_n\),满足关系\(\lg \left(S_{n}+1\right)=n\).求\(\{a_n\}\)的通项公式.
【解析】\(\because \lg \left(S_{n}+1\right)=n\),
\(\therefore S_{n}=10^{n}-1\)
当\(n≥2\)时,\(a_{n}=S_{n}-S_{n-1}=9 \times 10^{n-1}\)
当\(n=1\)时,\(a_1=S_1=9\)满足\(a_{n}=9 \times 10^{n-1}\),
\({\color{Red}{ (确定a_1是否满足上式)}}\)
\(\therefore a_{n}=9 \times 10^{n-1}\left(n \in N^{*}\right)\).
\({\color{Red}{(最后等式才由n≥2变成n∈N^*)}}\)

【典题2】已知数列\(\{a_n\}\)}的前\(n\)项和\(S_n\),\(a_1=1\),满足下列条件
①\(∀n∈N^*\),\(a_n>0\);②点\((a_n ,S_n)\)在函数\(f(x)=\dfrac{x^{2}+x}{2}\)的图象上;求数列\(\{a_n\}\)的通项\(a_n\)及前\(n\)项和\(S_n\).
【解析】由题意\(S_{n}=\dfrac{a_{n}^{2}+a_{n}}{2}\),
当\(n≥2\)时,\(a_{n}=S_{n}-S_{n-1}=\dfrac{a_{n}^{2}+a_{n}}{2}-\dfrac{a_{n-1}^{2}+a_{n-1}}{2}\),
整理,得\(\left(a_{n}+a_{n-1}\right)\left(a_{n}-a_{n-1}-1\right)=0\),
\({\color{Red}{(因式分解)}}\)
又\(∀n∈N^*\),\(a_n>0\),所以\(a_{n}+a_{n-1} \neq 0\)
即\(a_{n}-a_{n-1}=1\),
又\(a_1=1\),
\(∴\)数列\(\{a_n\}\)是首项为\(1\),公比为\(1\)的等差数列,
\(∴a_n=n\),\(S_{n}=\dfrac{n^{2}+n}{2}\).

【典题3】已知\(\{a_n\}\)中,\(a_1=1\),\(a_{n}=\dfrac{2 S_{n}^{2}}{2 S_{n}-1}\)\((n≥ 2)\),求\(a_n\).
【解析】当\(n≥2\)时,\(a_{n}=S_{n}-S_{n-1}\)
\(\therefore S_{n}-S_{n-1}=\dfrac{2 S_{n}^{2}}{2 S_{n}-1}\)
\(\therefore S_{n-1}-S_{n}=2 S_{n} S_{n-1}\)
两边同除以\(S_{n} S_{n-1}\),得\(\dfrac{1}{S_{n}}-\dfrac{1}{S_{n-1}}=2\)
\({\color{Red}{ (该变式技巧了解下)}}\)
\({\color{Red}{ (上两题是“消去”S_n得到数列\{a_n\}递推公式,该题“消去”a_n得到数列\{S_n\}的递推公式)}}\)
\(∴\)数列\(\left\{\dfrac{1}{S_{n}}\right\}\)为等差数列,公差为\(2\),首项为\(1\).
\(\therefore \dfrac{1}{S_{n}}=1+2(n-1)=2 n-1\),
解得\(S_{n}=\dfrac{1}{2 n-1}\),
\(\therefore a_{n}=S_{n}-S_{n-1}=\dfrac{1}{2 n-1}-\dfrac{1}{2 n-3}=\dfrac{-2}{(2 n-1)(2 n-3)}\)\((n≥ 2)\)
\({\color{Red}{(不要漏了大前提n≥ 2)}}\)
\(a_1=1\)不满足\(a_{n}=\dfrac{-2}{(2 n-1)(2 n-3)}\),
\(\therefore a_{n}= \begin{cases}1, & n=1 \\ \dfrac{-2}{(2 n-1)(2 n-3)}, & n \geq 2\end{cases}\).
【点拨】当题中得知\(S_n\)或\(a_n\)与\(S_n\)的关系式,则可利用公式\(a_{n}=\left\{\begin{array}{cc} S_{1} & n=1 \\ S_{n}-S_{n-1} & n \geq 2 \end{array}\right.\),消去\(a_n\)与\(S_n\),得到对应的递推公式进而求解\(a_n\),但最后都要注意确定\(a_1\)是否满足\(a_n=f(n)(n≥2)\).

巩固练习

1(★)已知数列\(\{a_n\}\)的前\(n\)项和\(S_n\)满足\(S_n=n^2+n-1\),求数列\(\{a_n\}\)的通项公式.

2(★★)已知无穷数列\(\{a_n\}\)的前\(n\)项和\(S_n\),并且\(a_n+S_n=1\),求\(\{a_n\}\)的通项公式.

3(★★)已知数列\(\{a_n\}\)的前\(n\)项和\(S_n\),满足\(a_2=-4\),\(2 S_{n}=n\left(a_{n}-7\right)\).求\(a_1\)和数列\(\{a_n\}\)的通项公式;

4(★★★)设数列\(\{a_n\}\)的前\(n\)项和\(S_n\),已知\(a_1=2\),\(a_2=8\),\(S_{n+1}+4 S_{n-1}=5 S_{n}(n \geq 2)\),求数列\(\{a_n\}\)的通项公式;

答案

1.\(a_{n}=\left\{\begin{array}{c} 1, n=1 \\ 2 n, n \geq 2 \end{array}\right.\)
2.\(a_{n}=\left(\dfrac{1}{2}\right)^{n}\)
3.\(a_1=-7\),\(a_n=3n-10(n∈N^*)\)
4.\(a_{n}=2^{2 n-1}\)

【方法三】累加法

\({\color{Red}{适用范围:}}\)递推式为\(a_{n+1}=a_{n}+f(n)\).
\({\color{Red}{方法:}}\)得到\(\rfloor a_{n+1}-a_{n}=f(n)\),利用累加的形式求出\(a_n\).

【典题1】已知数列\(\{a_n\}\)满足\(a_1=2\),\(a_{n+1}=a_{n}+\ln \left(1+\dfrac{1}{n}\right)\),求\(a_n\).
【解析】由条件知:\(a_{n+1}-a_{n}=\ln \left(1+\dfrac{1}{n}\right)=\ln \dfrac{n+1}{n}=\ln (n+1)-\ln n\)
\(∴n≥2\)时
\(\left\{\begin{aligned} a_{n}-a_{n-1} &=\ln n-\ln (n-1) \\ a_{n-1}-a_{n-2} &=\ln (n-1)-\ln (n-2) \\ a_{4}-a_{3} &=\ln 4-\ln 3 \\ a_{3}-a_{2} &=\ln 3-\ln 2 \\ a_{2}-a_{1} &=\ln 2-\ln 1 \end{aligned}\right.\),
把以上\(n-1\)个式子累加得\(a_{n}-a_{1}=\ln n-\ln 1=\ln n\),
\(\therefore a_{n}=a_{1}+\ln n=\ln n+2 \quad(n \geq 2)\),
\(a_1=2\)也满足\(a_{n}=\ln n+2\),
\(\therefore a_{n}=\ln n+2 \quad\left(n \in N^{*}\right)\).

【典题2】已知数列\(\{a_n\}\)满足\(a_{n+1}=a_{n}+2 \times 3^{n}+1\),\(a_1=3\),求数列\(\{a_n\}\)的通项公式.
【解析】由\(a_{n+1}=a_{n}+2 \times 3^{n}+1\)得\(a_{n+1}-a_{n}=2 \times 3^{n}+1\)
\(∴n≥2\)时,
\(\begin{aligned} &a_{n}=\left(a_{n}-a_{n-1}\right)+\left(a_{n-1}-a_{n-2}\right)+\cdots+\left(a_{3}-a_{2}\right)+\left(a_{2}-a_{1}\right)+a_{1} \\ &=\left(2 \times 3^{n-1}+1\right)+\left(2 \times 3^{n-2}+1\right)+\cdots+\left(2 \times 3^{2}+1\right)+\left(2 \times 3^{1}+1\right)+3 \\ &=2\left(3^{n-1}+3^{n-2}+\cdots+3^{2}+3^{1}\right)+(n-1)+3 \\ &=2 \dfrac{3\left(1-3^{n-1}\right)}{1-3}+(n-1)+3 \\ &=3^{n}+n-1 \end{aligned}\)
而\(a_1=3\)也满足\(a_n=3^n+n-1\),
\(∴a_n=3^n+n-1 (n∈N^*)\).

巩固练习

1(★)数列\(\{a_n\}\)满足\(a_1=3\),\(a_{n+1}-a_{n}=2 n-8\left(n \in N^{*}\right)\),则\(a_8=\)   .

2(★★)将正整数按一定的规则排成了如图所示的三角形数阵.根据这个排列规则,数阵中第\(20\)行从左至右的第\(3\)个数是   .
求数列的通项公式

3(★★)已知数列\(\{a_n\}\)满足\(a_{1}=\dfrac{1}{2}\),\(a_{n+1}=a_{n}+\dfrac{1}{n^{2}+n}\),求\(a_n\)

4(★★★)已知数列\(\{a_n\}\)的前\(n\)项和\(S_n\),\(S_n+a_n=n+2\),\(n∈N^*\).
(1)证明:数列\(\left\{a_{n}-1\right\}\)为等比数列;
(2)若数列\(\left\{b_{n}\right\}\)满足\(a_{n}=b_{n+1}-b_{n}+1\),\(b_1=1\),证明:\(b_n<2\).

答案

1.\(3\)
2.\(577\)
3.\(a_{n}=\dfrac{3}{2}-\dfrac{1}{n}\)
4.\((1)\)提示:定义法证明
\((2)\)提示:累加法求\(b_{n}=2-\dfrac{1}{2^{n}-1}\)

【方法四】累乘法

\({\color{Red}{适用范围:}}\)递推式为\(a_{n+1}=f(n) a_{n}\).
\({\color{Red}{方法:}}\)得到\(\dfrac{a_{n+1}}{a_{n}}=f(n)\),利用累乘的形式求出\(a_n\).

【典题1】已知\(\{a_n\}\)中,满足\(a_1=1\),\(a_{n}=a_{1}+2 a_{2}+3 a_{3}+\cdots+(n-1) a_{n-1}(n \geq 2)\),求\(a_n\).
【解析】由已知,得\(a_{n+1}=a_{1}+2 a_{2}+3 a_{3}+\cdots+(n-1) a_{n-1}+n a_{n}\),
用此式减去已知式,得
当\(n≥ 2\)时,\(a_{n+1}-a_{n}=n a_{n}\),
即\(a_{n+1}=(n+1) a_{n} \Rightarrow \dfrac{a_{n+1}}{a_{n}}=n+1\),
又\(a_2=a_1=1\),
\(\therefore\left\{\begin{array}{c} \dfrac{a_{n}}{a_{n-1}}=n \\ \dfrac{a_{n-1}}{a_{n-2}}=n-1 \\ \dfrac{a_{3}}{a_{2}}=3 \\ \dfrac{a_{2}}{a_{1}}=2 \end{array}\right.\),
将以上\(n-1\)个式子相乘,得\(\dfrac{a_{n}}{a_{1}}=\dfrac{n !}{2} (n≥ 2)\),
\(\therefore a_{n}=\dfrac{n !}{2} \quad(n \geq 2)\),
\(\therefore a_{n}= \begin{cases}1, & n=1 \\ \dfrac{n !}{2}, & n \geq 2\end{cases}\).
【典题2】设数列\(\{a_n\}\)是首项为1的正项数列,且\((n+1) a_{n+1}^{2}-n a_{n}^{2}+a_{n+1} a_{n}=0\), 求通项公式\(a_n\)
【解析】由\((n+1) a_{n+1}^{2}-n a_{n}^{2}+a_{n+1} a_{n}=0\),
可得\(n\left(a_{n+1}^{2}-a_{n}^{2}\right)+\left(a_{n+1}^{2}+a_{n+1} a_{n}\right)=0\),
即有\(n\left(a_{n+1}-a_{n}\right)\left(a_{n+1}+a_{n}\right)+a_{n+1}\left(a_{n+1}+a_{n}\right)=0\),
即有\(\left[(n+1) a_{n+1}-n a_{n}\right]\left(a_{n+1}+a_{n}\right)=0\)
由\(\{a_n\}\)是正项数列,可得\((n+1) a_{n+1}=n a_{n} \Rightarrow a_{n+1}=\dfrac{n}{n+1} a_{n}\),
则\(a_{n}=\dfrac{n-1}{n} a_{n-1}=\dfrac{n-1}{n} \cdot \dfrac{n-2}{n-1} a_{n-2}\)\(=\cdots=\dfrac{n-1}{n} \cdot \dfrac{n-2}{n-1} \cdots \dfrac{2}{3} \cdot \dfrac{1}{2} a_{1}\)\(=\dfrac{1}{n} a_{1}=\dfrac{1}{n}(n \geq 2)\),
\(a_1=1\)也满足\(a_{n}=\dfrac{1}{n}\),
\(∴a_{n}=\dfrac{1}{n},n∈N^*\).

巩固练习

1(★★)已知数列\(a_1\),\(\dfrac{a_{2}}{a_{1}}\),\(\dfrac{a_{3}}{a_{2}}\),⋯,\(\dfrac{a_{n}}{a_{n-1}}\)是首项为\(8\),公比为\(\dfrac{1}{2}\)的等比数列,则\(a_4\)等于  .

2(★★)在数列\(\{a_n\}\)中,\(a_1=a_2=1\),\(a_3=2\),且数列\(\left\{\dfrac{a_{n+1}}{a_{n}}\right\}\)为等比数列,则\(a_n=\)  .

3(★★)已知\(a_1=3\),\(a_{n+1}=\dfrac{3 n-1}{3 n+2} a_{n}(n \geq 1)\),求\(a_n\).

4(★★)已知\(a_1=1\),\(a_{n}=n\left(a_{n+1}-a_{n}\right)\left(n \in N^{*}\right)\),求数列\(\{a_n\}\)通项公式.

答案

1.\(64\)
2.\(2^{\dfrac{(n-2)(n-1)}{2}}\)
3.\(a_{n}=\dfrac{6}{3 n-1}\)
4.\(a_n=n\)

【方法五】构造法

对于一些不是等差等比数列的数列,求其通项公式,通过构造等差或等比数列来求其通项公式是一种很好的思路,其中的情况多样,方法有待定系数法、阶差法、取倒数法、取对数法等.我们要理解其中构造的技巧,做到举一反三.

情况1

递推公式为\(a_{n+1}=p a_{n}+q\)(\(p\),\(q\)为常数,\(p≠1\),\(pq≠0\))
\({\color{Red}{待定系数法:}}\)把原递推公式转化为:\(a_{n+1}+t=p\left(a_{n}+t\right)\),其中\(t=\dfrac{q}{p-1}\),再利用换元法转化为等比数列\(\left\{a_{n}+t\right\}\)求解;
\({\color{Red}{逐项相减法(阶差法):}}\)由\(a_{n+1}=p a_{n}+q\)得\(a_{n}=p a_{n-1}+q\),两式相减得\(a_{n+1}-a_{n}=p\left(a_{n}-a_{n-1}\right)\),即\(\left\{a_{n+1}-a_{n}\right\}\)是等比数列,再用累加法求解.

【典题1】已知数列\(\{a_n\}\)中,\(a_1=1\),\(a_{n+1}=2 a_{n}+3\), 求\(a_n\).
【解析】 \({\color{Red} {方法一\quad 待定系数法}}\)
设递推公式\(a_{n+1}=2 a_{n}+3\)可以转化为\(a_{n+1}+t=2\left(a_{n}+t\right)\)(\(t\)是个常数),
即\(a_{n+1}=2 a_{n}+t\),
与已知条件\(a_{n+1}=2 a_{n}+3\)比较可知\(t=3\),
\({\color{Red}{ (比较系数可求参数t )}}\)
故递推公式为\(a_{n+1}+3=2\left(a_{n}+3\right)\),
所以\(\left\{a_{n}+3\right\}\)是首项为\(a_1+3=4\),公比为\(2\)的等比数列,
\({\color{Red}{ (构造等比数列)}}\)
则\(a_{n}+3=4 \times 2^{n-1}=2^{n+1}\),
\(\therefore a_{n}=2^{n+1}-3\).

\({\color{Red}{方法二 \quad 逐项相减法}}\)
\(\because a_{n+1}=2 a_{n}+3\),\(\therefore a_{n}=2 a_{n-1}+3\)
两式相减得\(a_{n+1}-a_{n}=2\left(a_{n}-a_{n-1}\right)(n \geq 2)\)
\(∴\)数列\(\left\{a_{n+1}-a_{n}\right\}\)是以\(a_2-a_1=5-1=4\)为首项,公比\(q=2\)的等比数列, \({\color{Red}{(构造等比数列)}}\)
\(\therefore a_{n+1}-a_{n}=2^{n+1}\),
\({\color{Red}{(形如a_{n+1}-a_{n}=f(n)用累加法)}}\)
\(\therefore a_{n+1}=\left(a_{n+1}-a_{n}\right)+\left(a_{n}-a_{n-1}\right)+\cdots\left(a_{2}-a_{1}\right)+a_{1}\)\(=2^{n+1}+2^{n}+\cdots+4+1=2^{n+2}-3\)
\(\therefore a_{n}=2^{n+1}-3\).

情况2

递推公式为\(a_{n+1}=p a_{n}+k n+b\)(\(p\),\(k\),\(b\)为常数,\(p≠1\),\(pk≠0\))
\({\color{Red}{待定系数法:}}\) \(a_{n+1}=p a_{n}+k n+b\)可化为\(a_{n+1}+\lambda_{1}(n+1)+\lambda_{2}=A\left[a_{n}+\lambda_{1} n+\lambda_{2}\right]\)的形式,得到等比数列\(\left\{a_{n}+\lambda_{1} n+\lambda_{2}\right\}\)求出\(a_n\).
\({\color{Red}{逐项相减法(阶差法):}}\)由\(a_{n+1}=p a_{n}+k n+b\)得\(a_{n}=p a_{n-1}+k(n-1)+b\),两式相减得\(a_{n+1}-a_{n}=p\left(a_{n}-a_{n-1}\right)+k\),即令\(b_{n}=a_{n+1}-a_{n}\)得\(b_{n}=p b_{n-1}+k\),再用递推公式形如\(a_{n+1}=p a_{n}+q\)的方法求解.

【典题1】设数列\(\{a_n\}\):\(a_1=4\),\(a_{n}=3 a_{n-1}+2 n-1(n \geq 2)\),求\(a_n\).
【解析】 \({\color{Red}{方法一 \quad 待定系数法}}\)
令\(a_{n}+\lambda_{1} n+\lambda_{2}=3\left[a_{n-1}+\lambda_{1}(n-1)+\lambda_{2}\right]\)
化简得:\(a_{n}=3 a_{n-1}+2 \lambda_{1} n+2 \lambda_{2}-3 \lambda_{1}\)
与已知条件\(a_{n}=3 a_{n-1}+2 n-1\)比较系数,
所以\(\left\{\begin{array}{c} 2 \lambda_{1}=2 \\ 2 \lambda_{2}-3 \lambda_{1}=-1 \end{array}\right.\), 解得\(\left\{\begin{array}{l} \lambda_{1}=1 \\ \lambda_{2}=1 \end{array}\right.\),
\({\color{Red}{ (比较系数可求参数λ_1,λ_2)}}\)
所以\(a_{n}+n+1=3\left(a_{n-1}+n\right)\)
\(∴\)数列\(\left\{a_{n}+n+1\right\}\)是以\(a_1+2=6\)为首项,\(3\)为公比的等比数列.
\(\therefore a_{n}+n+1=6 \times 3^{n-1}=2 \times 3^{n}\),
\(∴a_n=2×3^n-n-1\).
\({\color{Red}{方法二 \quad 逐项相减法(阶差法)}}\)
\(\because a_{n}=3 a_{n-1}+2 n-1(n \geq 2)\)①
\(\therefore a_{n+1}=3 a_{n}+2 n+1\)②
由②-①得\(a_{n+1}-a_{n}=3\left(a_{n}-a_{n-1}\right)+2\)
令\(b_{n}=a_{n+1}-a_{n}\),则\(b_{n}=3 b_{n-1}+2\)
\({\color{Red}{ (回归递推公式形如a_{n}=p a_{n-1}+q的形式)}}\)
\(\therefore b_{n}+1=3\left(b_{n-1}+1\right)\)
\(∴\)数列\(\left\{b_{n}+1\right\}\)是以首项\(b_1+1=a_2-a_1+1=15-4+1=12\),公比\(q=3\)的等比数列,
则\(b_{n}+1=12 \times 3^{n-1}=4 \times 3^{n} \Rightarrow b_{n}=4 \times 3^{n}-1\)
\(\therefore a_{n+1}=\left(a_{n+1}-a_{n}\right)+\left(a_{n}-a_{n-1}\right)+\cdots\left(a_{2}-a_{1}\right)+a_{1}\) \({\color{Red}{(累加法)}}\)
\(=\left(4 \times 3^{n}-1\right)+\left(4 \times 3^{n-1}-1\right)+\cdots+\left(4 \times 3^{1}-1\right)+4\)
\(=2 \times 3^{n+1}-n-2\)
\(∴a_n=2×3^n-n-1\).
【点拨】二种方法比较还是待定系数法过程显得简洁些,形如\(a_{n+1}=p a_{n}+q\)和\(a_{n+1}=p a_{n}+k n+b\)都可用待定系数法构造等比数列来求\(a_n\),那是否形如\(a_{n+1}=p a_{n}+f(n)\)都可行呢?那形如“\(a_{n+1}=p a_{n}+\)二次函数”如何?试试题目:已知\(a_1=6\),\(a_{n+1}=2 a_{n}-n^{2}+n+2\),求\(a_n\).(答案:\(a_{n}=2^{n+1}+n^{2}+n\)).

情况3

递推公式为\(a_{n}=\dfrac{p a_{n-1}}{q a_{n-1}+t}\)(\(p\),\(q\),\(t\)为常数)
\({\color{Red}{取倒数法:}}\)递推公式两边取倒数,\(\dfrac{1}{a_{n}}=\dfrac{q a_{n-1}+t}{p a_{n-1}}=\dfrac{q}{p}+\dfrac{t}{p} \cdot \dfrac{1}{a_{n-1}}\),令\(b_{n}=\dfrac{1}{a_{n}}\),若\(p=t\),则问题\(b_n\)是等差数列;若\(p≠t\),问题转化为递推公式为:\(a_{n+1}=p_{1} a_{n}+q_{1}\)的方法处理.

【典题1】已知数列\(\{a_n\}\)中,\(a_1=1\),\(a_{n}=\dfrac{a_{n-1}}{5 a_{n-1}+2}(n \geq 2)\),求通项公式\(a_n\).
【解析】对\(a_{n}=\dfrac{a_{n-1}}{5 a_{n-1}+2}\)
两边取倒数,得\(\dfrac{1}{a_{n}}=\dfrac{5 a_{n-1}+2}{a_{n-1}}=2 \cdot \dfrac{1}{a_{n-1}}+5\)
令\(b_{n}=\dfrac{1}{a_{n}}\),则\(b_{n}=2 b_{n-1}+5\)
\(\therefore b_{n}+5=2\left(b_{n-1}+5\right)\)
\(∴\)数列\(\left\{b_{n}+5\right\}\)是首项为\(b_{1}+5=\dfrac{1}{a_{1}}+5=6\),公比为\(2\)的等比数列, \({\color{Red}{ (构造了等比数列)}}\)
\(\therefore b_{n}+5=6 \times 2^{n-1}=3 \times 2^{n} \Rightarrow b_{n}=3 \times 2^{n}-5\)
\(\therefore a_{n}=\dfrac{1}{b_{n}}=\dfrac{1}{3 \times 2^{n}-5}\).

【典题2】已知\(\{a_n\}\)中,\(a_1=1\),\(S_n\)是数列的前\(n\)项和,且\(S_{n+1}=\dfrac{S_{n}}{4 S_{n}+3}(n \geq 1)\),求\(a_n\).
【解析】递推式\(S_{n+1}=\dfrac{S_{n}}{4 S_{n}+3}\)
\({\color{Red}{ (可利用取倒数法求出数列S_n的通项S_n,再用a_n与S_n的关系式求a_n)}}\)
两边取倒数可变形为\(\dfrac{1}{S_{n+1}}=3 \cdot \dfrac{1}{S_{n}}+4\)(1)
则有\(\dfrac{1}{s_{n+1}}+2=3\left(\dfrac{1}{s_{n}}+2\right)\).
故数列\(\left\{\dfrac{1}{S_{n}}+2\right\}\)是以\(\dfrac{1}{S_{1}}+2=3\)为首项,\(3\)为公比的等比数列.
\(\therefore \dfrac{1}{S_{n}}+2=3 \cdot 3^{n-1}=3^{n}\),\(\therefore S_{n}=\dfrac{1}{3^{n}-2}\).
当\(n≥ 2\),\(a_{n}=S_{n}-S_{n-1}\)\(=\dfrac{1}{3^{n}-2}-\dfrac{1}{3^{n-1}-2}=\dfrac{-2 \cdot 3^{n}}{3^{2 n}-8 \cdot 3^{n}+12}\).
所以数列\(\{a_n\}\)的通项公式是\(a_{n}= \begin{cases}1, & (n=1) \\ \dfrac{-2 \cdot 3^{n}}{3^{2 n}-8 \cdot 3^{n}+12}, & (n \geq 2)\end{cases}\).

情况4

递推公式为\(a_{n+1}=p a_{n}+q^{n}\)(其中\(p\),\(q\)均为常数,\(pq(p-1)(q-1)≠0\)). (或\(a_{n+1}=p a_{n}+r q^{n}\),其中\(p\),\(q\),\(r\)均为常数) .
\({\color{Red}{方法一}}\) 在原递推公式两边同除以\(q^{n+1}\),得\(\dfrac{a_{n+1}}{q^{n+1}}=\dfrac{p}{q} \cdot \dfrac{a_{n}}{q^{n}}+\dfrac{1}{q}\),令\(b_{n}=\dfrac{a_{n}}{q^{n}}\),得\(b_{n+1}=\dfrac{p}{q} b_{n}+\dfrac{1}{q}\)再待定系数法解决;
\({\color{Red}{方法二}}\) 在原递推公式两边同除以\(p^{n+1}\),得\(\dfrac{a_{n+1}}{p^{n+1}}=\dfrac{a_{n}}{p^{n}}+\dfrac{1}{p} \cdot\left(\dfrac{q}{p}\right)^{n}\), 令\(b_{n}=\dfrac{a_{n}}{p^{n}}\),得\(b_{n+1}=b_{n}+\dfrac{1}{p} \cdot\left(\dfrac{q}{p}\right)^{n}\)再用累加法求解;
\({\color{Red}{方法三}}\) 待定系数法 设\(a_{n+1}+\lambda q^{n+1}=p\left(a_{n}+\lambda q^{n}\right)\),通过比较系数,求出\(λ\),构造出等比数列\(\left\{a_{n}+\lambda q^{n}\right\}\)再求解,此时要求\(p≠q\),否则该法失效.

【典题1】已知数列\(\{a_n\}\)满足\(a_{n+1}=2 a_{n}+4 \times 3^{n}\),\(a_1=9\),求数列\(\{a_n\}\)的通项公式.
【解析】 \({\color{Red}{方法一\quad (两边同除以3^{n+1) }}\)
\(a_{n+1}=2 a_{n}+4 \times 3^{n}\)两边同除以\(3^{n+1}\),得\(\dfrac{a_{n+1}}{3^{n+1}}=\dfrac{2}{3} \cdot \dfrac{a_{n}}{3^{n}}+\dfrac{4}{3}\),
\({\color{Red}{(转化为递推公式为a_{n+1}=p a_{n}+q的情况)}}\)
令\(b_{n}=\dfrac{a_{n}}{3^{n}}\),则\(b_{n+1}=\dfrac{2}{3} b_{n}+\dfrac{4}{3}\),
\(\therefore b_{n+1}-4=\dfrac{2}{3}\left(b_{n}-4\right)\)
\(\therefore b_{n}-4=\left(b_{1}-4\right) \cdot\left(\dfrac{2}{3}\right)^{n-1}=-\left(\dfrac{2}{3}\right)^{n-1}\)\(\Rightarrow b_{n}=4-\left(\dfrac{2}{3}\right)^{n-1}\)
\(\therefore a_{n}=4 \cdot 3^{n}-3 \cdot 2^{n-1}\).
\({\color{Red}{方法二\quad (两边同除以2^{n+1}) }}\)
\(a_{n+1}=2 a_{n}+4 \times 3^{n}\)两边同除以\(2^{n+1}\),得\(\dfrac{a_{n+1}}{2^{n+1}}=\dfrac{a_{n}}{2^{n}}+2 \cdot\left(\dfrac{3}{2}\right)^{n}\),
\(\therefore b_{n}=\left(b_{n}-b_{n-1}\right)+\left(b_{n-1}-b_{n-2}\right)+\cdots\left(b_{2}-b_{1}\right)+b_{1}\) \({\color{Red}{(累加法)}}\)
\(=2 \cdot\left(\dfrac{3}{2}\right)^{n-1}+2 \cdot\left(\dfrac{3}{2}\right)^{n-2}+\cdots+2 \cdot\left(\dfrac{3}{2}\right)^{1}+\dfrac{9}{2}=6 \cdot\left(\dfrac{3}{2}\right)^{n-1}-\dfrac{3}{2}(n \geq 2)\)
\(\because b_{1}=\dfrac{9}{2}\)也满足上式
\(\therefore b_{n}=6 \cdot\left(\dfrac{3}{2}\right)^{n-1}-\dfrac{3}{2}(n \in N)\),
\(\therefore a_{n}=2^{n} \cdot b_{n}=4 \cdot 3^{n}-3 \cdot 2^{n-1}\).
\({\color{Red}{方法三\quad 待定系数法}}\)
设\(a_{n+1}+\lambda \cdot 3^{n+1}=2\left(a_{n}+\lambda \cdot 3^{n}\right)\),
化简得\(a_{n+1}=2 a_{n}-\lambda \cdot 3^{n}\),
与已知条件\(a_{n+1}=2 a_{n}+4 \times 3^{n}\)比较可知\(λ=-4\),
则\(a_{n+1}-4 \cdot 3^{n+1}=2\left(a_{n}-4 \cdot 3^{n}\right)\)
\(\therefore a_{n}-4 \cdot 3^{n}=\left(a_{1}-12\right) \cdot 2^{n-1}=-3 \cdot 2^{n-1}\),
\(\therefore a_{n}=4 \cdot 3^{n}-3 \cdot 2^{n-1}\).
【点拨】方法技巧主要都是体现在通过构造等差等比数列和把问题转化为前面“已知模型”去.

情况5

递推公式为\(a_{n+1}=p a_{n}^{r} \text { 开 }\)型
\({\color{Red}{方法一\quad 对数变换法:}}\)该类型是等式两边取对数后转化为前边的类型,然后再用递推法或待定系法构造等比数列求出通项.
两边取对数得
\(\lg a_{n+1}=\lg \left(p a_{n}^{r}\right) \Rightarrow \lg a_{n+1}=\lg p+\operatorname{rlg} a_{n}\)
设\(b_{n}=\lg a_{n}\)
\(∴\)原等式变为\(b_{n+1}=r b_{n}+\lg p\)即变为基本型.
\({\color{Red}{方法二\quad 迭代法}}\),反复迭代使用\(a_{n+1}=p a_{n}^{r}\),一直推到\(a_1\),
\(a_{n}=p a_{n-1}^{r}=p\left(p a_{n-2}^{r}\right)^{r}=p^{r+1} a_{n-2}^{r^{2}}\)\(=p^{r+1}\left(p a_{n-3}^{r}\right)^{r^{2}}=p^{r^{2}+r+1} a_{n-3}^{r^{3}}=\cdots=f(r) a_{1}^{r^{n-1}}\).


【典题1】设正项数列\(\{a_n\}\)满足\(a_1=1\),\(a_{n}=3 a_{n-1}^{2}(n \geq 2)\), 求通项公式\(a_n\).
【解析】 \({\color{Red}{方法一 \quad 对数变换法}}\)
\(a_{n}=3 a_{n-1}^{2}(n \geq 2)\)
将等式两边取对数得\(\log _{3} a_{n}=1+2 \log _{3} a_{n-1}\)\(\Rightarrow \log _{3} a_{n}+1=2\left(\log _{3} a_{n-1}+1\right)\)
\({\color{Red}{(为了计算简便取对数log_3x})}}\)
则\(\left\{\log _{3} a_{n}+1\right\}\)是以\(\log _{3} a_{1}+1=1\)为首项,公比为\(2\)等比数列,
\(\therefore \log _{3} a_{n}+1=2^{n-1}\),
\(\therefore a_{n}=3^{2^{n-1}-1}\).
\({\color{Red}方法二 \quad 迭代法{}}\)
由\(a_{n}=3 a_{n-1}^{2}(n \geq 2)\)可迭代得
\(a_{n}=3 a_{n-1}^{2}=3\left(3 a_{n-2}^{2}\right)^{2}=3 \cdot 3^{2} a_{n-2}^{2^{2}}\)\(=3 \cdot 3^{2}\left(3 a_{n-3}^{2}\right)^{2^{2}}=3 \cdot 3^{2} \cdot 3^{4} a_{n-3}^{2^{3}}=\cdots\)\(=3 \cdot 3^{2} \cdot 3^{4} \cdot 3^{2^{n-2}} a_{1}^{2^{n-1}}=3^{2^{n-1}-1}\)
\({\color{Red}{(在迭代的过程中,逐一保持“原始数值”,找到数值变化的规律,比如指数与下标的关系之类的) }}\)

巩固练习

1(★★)数列\(\{a_n\}\)中,\(a_1=1\),\(a_{n+1}=\dfrac{2 a_{n}}{a_{n}+2}\left(n \in \boldsymbol{N}^{*}\right)\),则\(\dfrac{2}{101}\)是这个数列的第 项.

2(★★)若数列\(\{a_n\}\)中,\(a_1=3\)且\(a_{n+1}=a_{n}^{2}\)(\(n\)是正整数),则它的通项公式是\(a_n=\).

3(★★)已知数列\(\{a_n\}\)满足\(a_{1}=\dfrac{5}{2}\),\(a_{n}=3 a_{n-1}+1\)\(\left(n \geq 2, n \in \boldsymbol{N}^{*}\right)\).求数列\(\{a_n\}\)的通项公式.

4(★★)已知数列\(\{a_n\}\)中,当\(n≥2\)时,\(a_{n}=\dfrac{a_{n-1}}{3 a_{n-1}+1}\),\(a_1=1\),求通项公式\(a_n\).

5(★★)已知数列\(\{a_n\}\)满足\(a_{1}=\dfrac{7}{3}\),\(a_{n+1}=3 a_{n}-4 n+2\).求通项公式\(a_n\).

6(★★)已知数列\(\{a_n\}\)中,\(a_{1}=\dfrac{5}{6}\),\(a_{n+1}=\dfrac{1}{3} a_{n}+\left(\dfrac{1}{2}\right)^{n+1}\),求\(a_n\).

7(★★)数列\(\{a_n\}\)前\(n\)项和\(S_{n}=4-a_{n}-\dfrac{1}{2^{n-2}}\),求通项公式\(a_n\).

8(★★★★)已知数列\(\{a_n\}\)的各项都是正数,且满足\(a_0=1\),\(a_{n+1}=\dfrac{1}{2} a_{n}\left(4-a_{n}\right)\),\(n∈N\).求数列\(\{a_n\}\)的通项公式\(a_n\).

答案

1.\(100\)
2.\(a_{n}=3^{2^{n-1}}\)
3.\(a_{n}=3^{n}-\dfrac{1}{2}\)
4.\(a_{n}=\dfrac{1}{3 n-2}\)
5.\(a_{n}=3^{n-2}+2 n\)
6.\(a_{n}=\dfrac{3}{2^{n}}-\dfrac{2}{3^{n}}\)
7.\(a_{n}=\dfrac{n}{2^{n-1}}\)
8.\(a_{n}=2-\left(\dfrac{1}{2}\right)^{2^{n}-1}\)

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