数列专题 1 求数列的通项公式
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选择性第二册同步巩固,难度3颗星!
基础知识
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.
基本方法
方法一 an与Sn的关系公式法
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.
方法 利用\(a_n\)与\(S_n\)的关系\(a_n= \begin{cases}S_1 & n=1 \\ S_n-S_{n-1} & n \geq 2\end{cases}\),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.
解析 \(\because a_n+S_n=1\),当\(n=1\)时,\(a_1=\dfrac{1}{2}\),
\(\because a_{n+1}=S_{n+1}-S_n=a_n-a_{n+1}\),\(\therefore a_{n+1}=\dfrac{1}{2} a_n\),
又\(a_1=\dfrac{1}{2}≠0\),
\(\therefore \{a_n\}\)是以首项为\(\dfrac{1}{2}\),公比为\(\dfrac{1}{2}\)的等比数列,
\(\therefore a_n=\left(\dfrac{1}{2}\right)^n\).
【典题2】 \(S_n\)为数列\(\{a_n\}\)的前\(n\)项和,\(a_1=1\),\(S_n=na_n-n^2+n\).求\(\{a_n\}\)的通项公式.
解析 由已知\(S_n=na_n-n^2+n\) ①,
\(\therefore S_{n-1}=(n-1)a_{n-1}-(n-1)^2+n-1\left(n≥2\right)\)②,
由①-②,得\(a_n=na_n-(n-1)a_{n-1}-2(n-1)\),
即\((n-1)a_n-(n-1)a_{n-1}=2(n-1)\),
\(\therefore a_n-a_{n-1}=2\),\(n≥2\)且\(n∈N^*\),
\(\therefore \{a_n\}\)是以\(2\)为公差的等差数列,
\(\therefore a_n=a_1+(n-1)d=1+2(n-1)=2n-1\).
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.已知数列\(\{a_n\}\)的前\(n\)项和为\(S_n\),若\(a_1=1\),\(S_n=a_{n+1}-2\).求数列\(\{a_n\}\)的通项公式.
参考答案
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答案 \(a_n=\left\{\begin{array}{c} 1, n=1 \\ 2 n, n \geq 2 \end{array}\right.\)
解析 当\(n≥2\)时,\(a_n=S_n-S_{n-1}=2n\)
当\(n=1\)时,\(a_1=S_1=1\)不满足\(a_n=2n\),
\(\therefore a_n=\left\{\begin{array}{c} 1, n=1 \\ 2 n, n \geq 2 \end{array}\right.\). -
答案 \(a_n=\left\{\begin{array}{l} 1, n=1 \\ 3 \cdot 2^{n-2}, n \geq 2 \end{array}\right.\)
解析 当\(n≥2\)时,由 \(\left\{\begin{array}{l} S_n=a_n+1-2 \\ S_{n-1}=a_n-2 \end{array}\right.\),得\(a_n=a_{n+1}-a_n\),
即\(a_{n+1}=2a_n\),故数列\(\{a_n\}\)从第二项起为等比数列,
\(S_1=a_2-2\),即\(a_2=3\),
故当 \(n≥2\)时,\(a_n=3⋅2^{n-2}\),
故\(a_n=\left\{\begin{array}{l} 1, n=1 \\ 3 \cdot 2^{n-2}, n \geq 2 \end{array}\right.\).
方法二 累加法
适用范围 递推式为 \(a_{n+1}=a_n+f\left(n\right)\).
方法 得到\(a_{n+1}-a_n=f\left(n\right)\),利用累加的形式求出\(a_n\).
【典题1】数列\(\{a_n\}\)满足\(a_1=3\),\(a_{n+1}-a_n=2n-8\left(n∈N^*\right)\),则\(a_8=\) .
解析 数列\(\{a_n\}\)中,\(a_1=3\),\(a_{n+1}-a_n=2n-8\left(n∈N^*\right)\),
\(\therefore a_n-a_{n-1}=2n-10\),
\(a_{n-1}-a_{n-2}=2n-12\),
\(…\)
\(a_3-a_2=-4\),
\(a_2-a_1=-6\),
\(\therefore a_n-a_1=-6-4-\cdots+(2 n-12)+(2 n-10)\)\(=\dfrac{(n-1)[-6+(2 n-10)]}{2}=(n-1)(n-8)\),
\(\therefore a_n=(n-1)\left(n-8\right)+3\),
\(\therefore a_8=3\).
【典题2】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: \(a_{n+1}-a_n=\ln \left(1+\dfrac{1}{n}\right)=\ln \dfrac{n+1}{n}=\ln (n+1)-\ln n\)
\(\therefore n≥2\)时
\(\left\{\begin{aligned}
a_n-a_{n-1}=& \ln n-\ln (n-1) \\
a_{n-1}-a_{n-2} &=\ln (n-1)-\ln (n-2) \\
& \ldots \ldots \\
a_4-a_3 &=\ln 4-\ln 3 \\
a_3-a_2 &=\ln 3-\ln 2 \\
a_2-a_1 &=\ln 2-\ln 1
\end{aligned}\right.\) ,
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\(a_n=\ln n+2\),
\(\therefore a_n=\ln n+2 \left(n∈N^*\right)\).
巩固练习
1.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 \(\underline{\quad \quad}\) .
2.已知数列\(\{a_n\}\)满足\(a_{n+1}=a_n+2× 3^n+1\) , \(a_1=3\),求数列\(\{a_n\}\)的通项公式.
参考答案
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答案 \(577\)
解析 设各行的首项组成数列\(\{a_n\}\),则\(a_2-a_1=3\),\(a_3-a_2=6\),…,\(a_n-a_{n-1}=3(n-1)\)
叠加可得: \(a_n-a_1=3+6+\cdots+3(n-1)=\dfrac{3 n(n-1)}{2}\),
\(\therefore a_n=\dfrac{3 n(n-1)}{2}+1\),
\(\therefore a_{20}=\dfrac{3 \times 20 \times 19}{2}+1=571\),
\(\therefore\) 数阵中第\(20\)行从左至右的第\(3\)个数是\(571+2×3=577\),
故答案:\(577\). -
答案 \(a_n=3^n+n-1 \left(n∈N^*\right)\).
解析 由\(a_{n+1}=a_n+2× 3^n+1\)得\(a_{n+1}-a_n=2×3^n+1\)
\(\therefore n≥2\)时,
\(a_n=\left(a_n-a_{n-1} \right)+\left(a_{n-1}-a_{n-2} \right)+⋯+\left(a_3-a_2\right)+\left(a_2-a_1\right)+a_1\)
\(=\left(2× 3^{n-1}+1\right)+\left(2× 3^{n-2}+1\right)+⋯+\left(2× 3^2+1\right)+\left(2× 3^1+1\right)+3\)
\(=2\left(3^{n-1}+3^{n-2}+⋯+3^2+3^1\right)+{n-1}+3\)
\(=2\cdot \dfrac{3\left(1-3^{n-1}\right)}{1-3}+(n-1)+3\)
\(=3^n+n-1\)
而\(a_1=3\)也满足\(a_n=3^n+n-1\),
\(\therefore a_n=3^n+n-1 \left(n∈N^*\right)\).
方法三 累乘法
适用范围 递推式为 \(a_{n+1}=f\left(n\right) a_n\).
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\(a_1=1\), \(a_n=a_1+2a_2+3a_3+⋯+(n-1)a_{n-1} \left(n≥ 2\right)\),求\(a_n\).
解析 由已知,得\(a_{n+1}=a_1+2a_2+3a_3+⋯+(n-1)a_{n-1} +na_n\),
用此式减去已知式,得
当\(n≥ 2\)时,\(a_{n+1}-a_n=na_n\),
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\(\dfrac{a_n}{a_1}=\dfrac{n !}{2} \quad(n \geq 2)\),
\(\therefore\dfrac{a_n}{a_1}=\dfrac{n !}{2} \quad(n \geq 2)\),
\(\therefore a_n=\left\{\begin{array}{l}
1, \quad n=1 \\
\dfrac{n !}{2}, n \geq 2
\end{array}\right.\).
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( )
A.\(8\) \(\qquad \qquad \qquad \qquad\) B.\(32\) \(\qquad \qquad \qquad \qquad\) C.\(64\) \(\qquad \qquad \qquad \qquad\) D.\(128\)
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,求\(a_n\).
参考答案
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答案 \(C\)
解析 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 \(\dfrac{a_4}{a_3}=8 \times\left(\dfrac{1}{2}\right)^3=1\).
\(a_4=a_1 \times \dfrac{a_2}{a_1} \times \dfrac{a_3}{a_2} \times \dfrac{a_4}{a_3}=8 \times 4 \times 2 \times 1=64\).
故选:\(C\). -
答案 \(2^{\dfrac{(n-2)(n-1)}{2}}\)
解析 由题意可得, \(\dfrac{a_2}{a_1}=1\), \(\dfrac{a_3}{a_2}=2\),
故数列 \(\left\{\dfrac{a_{n+1}}{a_n}\right\}\)是以\(1\)为首项,以\(2\)为公比的等比数列,
所以 \(\dfrac{a_{n+1}}{a_n}=2^{n-1}\),即 \(\dfrac{a_n}{a_{n-1}}=2^{n-1} \times \dfrac{1}{2}=2^{n-2}\),
\(\dfrac{a_2}{a_1}=1\), \(\dfrac{a_3}{a_2}=2\),\(…\) ,\(\dfrac{a_n}{a_{n-1}}=2^{n-2}\)
以上\(n-1\)个式子相乘可得,
\(\dfrac{a_n}{a_1}=1 \times 2 \times 2^2 \times \ldots \times 2^{n-2}=2^{1+2+\cdots+n-2}=2^{\frac{(n-2)(n-1)}{2}}\).
故答案为: \(2^{\frac{(n-2)(n-1)}{2}}\).
方法四 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\(a_{n+1}=pa_n+q\) \((p,q\)为常数,\(p≠1\),\(pq≠0 )\)
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\(\{a_n+t\}\)求解;
【典题1】 已知数列\(\{a_n\}\)中,\(a_1=1\), \(a_{n+1}=2a_n+3\), 求\(a_n\).
解析 设递推公式\(a_{n+1}=2a_n+3\)可以转化为\(a_{n+1}+t=2\left(a_n+t\right)\) \((t\)是个常数\()\),
即\(a_{n+1}=2a_n+t\),
与已知条件\(a_{n+1}=2a_n+3\)比较可知\(t=3\),(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\(\{a_n+3\}\)是首项为\(a_1+3=4\),公比为\(2\)的等比数列, (构造等比数列)
则\(a_n+3=4× 2^{n-1}=2^{n+1}\),
\(\therefore a_n=2^{n+1}-3\).
情况2 递推公式为\(a_{n+1}=pa_n+q^n\)\((\)其中\(p,q\)均为常数,\(pq\left(p-1\right)\left(q-1\right)≠0)\).
(或\(a_{n+1}=pa_n+rq^n\),其中\(p\),\(q\) , \(r\)均为常数) .
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;
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.
【典题1】已知数列\(\{a_n \}\)满足\(a_{n+1}=2a_n+4×3^n\) ,\(a_1=9\),求数列\(\{a_n\}\)的通项公式.
解析 方法一 (两边同除以\(3^{n+1}\))
\(a_{n+1}=2a_n+4×3^n\)两边同除以\(3^{n+1}\),得\(\dfrac{a_{n+1}}{3^{n+1}}=\dfrac{2}{3} \cdot \dfrac{a_n}{3^n}+\dfrac{4}{3}\),
(转化为递推公式为\(a_{n+1}=pa_n+q\)的情况)
令 \(b_n=\dfrac{a_n}{3^n}\),则 \(b_{n+1}=\dfrac{2}{3} b_n+\dfrac{4}{3} \text {, }\),
\(\therefore b_{n+1}-4=\dfrac{2}{3}\left(b_n-4\right)\),
\(\therefore b_n-4=\left(b_1-4\right) \cdot\left(\dfrac{2}{3}\right)^{n-1}=-\left(\dfrac{2}{3}\right)^{n-1} \Rightarrow b_n=4-\left(\dfrac{2}{3}\right)^{n-1}\),
\(\therefore a_n=4 \cdot 3^n-3 \cdot 2^{n-1}\).
方法二 (两边同除以\(2^{n+1}\))
\(a_{n+1}=2a_n+4×3^n\)两边同除以\(2^{n+1}\),得 \(\dfrac{a_{n+1}}{2^{n+1}}=\dfrac{a_n}{2^n}+2 \cdot\left(\dfrac{3}{2}\right)^n\),
令 \(b_n=\dfrac{a_n}{2^n}\),则 \(b_{n+1}=b_n+2 \cdot\left(\dfrac{3}{2}\right)^n\),
\(\therefore b_n=\left(b_n-b_{n-1} \right)+\left(b_{n-1}-b_{n-2} \right)+⋯\left(b_2-b_1 \right)+b_1\),
\(=2\cdot \left(\dfrac{3}{2}\right)^{n-1}+\cdot \left(\dfrac{3}{2}\right)^{n-2}+⋯+2\cdot \left(\dfrac{3}{2}\right)^1+ \dfrac{9}{2}=6\cdot \left(\dfrac{3}{2}\right)^{n-1}-\dfrac{3}{2} \left(n≥2\right)\),
\(\because b_1= \dfrac{9}{2}\)也满足上式,\(\therefore b_n=6\cdot \left(\dfrac{3}{2}\right)^{n-1}-\dfrac{3}{2} \left(n∈N\right)\),
\(\therefore a_n=2^n\cdot b_n=4\cdot 3^n-3\cdot 2^{n-1}\).
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)
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 \(a_n=\dfrac{a_{n-1}}{5 a_{n-1}+2}(n \geq 2)\),
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\(b_n=2b_{n-1}+5\),
\(\therefore b_n+5=2\left(b_{n-1}+5\right)\),
\(\therefore\) 数列\(\{b_n+5\}\)是首项为\(b_1+5=\dfrac{1}{a_1}+5=6\),公比为\(2\)的等比数列,
\(\therefore b_n+5=6×2^{n-1}= 3×2^n⇒b_n=3×2^n-5\),
\(\therefore a_n=\dfrac{1}{b_n}=\dfrac{1}{3 \times 2^n-5}\).
巩固练习
1.已知数列\(\{a_n\}\)满足\(a_1=\dfrac{5}{2}\),\(a_n=3a_{n-1}+1\left(n≥2,n∈N^*\right)\).求数列\(\{a_n\}\)的通项公式.
2.已知数列\(\{a_n\}\)中,\(a_1=1\), \(a_{n+1}=\dfrac{a_n}{3+4 a_n}\),求\(\{a_n\}\)的通项公式\(a_n\).
3.已知数列\(\{a_n\}\)的前\(n\)项和为\(S_n\),且\(a_1=2\),\(S_{n+1}=3S_n+2\).
(1) 证明:\(\{S_n+1\}\)是等比数列;
(2) 求数列\(\{a_n\}\)的通项公式.
4.数列\(\{a_n\}\)前\(n\)项和 \(S_n=4-a_n-\dfrac{1}{2^{n-2}}\) .求通项公式\(a_n\).
5.已知各项都为正数的数列\(\{a_n\}\)满足\(a_{n+2}=2a_{n+1}+3a_n\) .
(1) 证明: 数列\(\{a_n+a_{n+1} \}\)为等比数列,
(2) 若\(a_1=\dfrac{1}{2}\),\(a_2=\dfrac{3}{2}\),求\(\{a_n\}\)的通项公式.
参考答案
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答案 \(a_n=3^n-\dfrac{1}{2}\).
解析 数列\(\{a_n\}\)满足\(a_1=\dfrac{5}{2}\),\(a_n=3a_{n-1}+1\left(n≥2,n∈N^*\right)\),
整理得\(a_n+\dfrac{1}{2}=3\left(a_{n-1}+\dfrac{1}{2}\right)\),
设\(b_n=a_n+\dfrac{1}{2}\),
所以数列列\(\{b_n\}\)为是以\(3\)为首项,\(3\)为公比的等比数列.
\(\therefore b_n=3^n\) ,则\(a_n=b_n-\dfrac{1}{2}=3^n-\dfrac{1}{2}\). -
答案 \(a_n=\dfrac{1}{3^n-2}\)
解析 由题意得 \(a_{n+1}=\dfrac{a_n}{3+4 a_n}\),
两边取倒数得,\(\dfrac{1}{a_{n+1}}=\dfrac{3+4 a_n}{a_n}=\dfrac{3}{a_n}+4\),
\(\therefore \dfrac{1}{a_{n+1}}+2=3\left(\dfrac{1}{a_n}+2\right)\),且 \(\dfrac{1}{a_1}+2=3\),
\(\therefore\) 数列\(\left\{\dfrac{1}{a_n}+2\right\}\)是以\(3\)为首项,\(3\)为公比的等比数列,
\(\therefore \dfrac{1}{a_n}+2=3^n\),\(\therefore a_n=\dfrac{1}{3^n-2}\). -
答案 (1) 略;(2) \(a_n=2⋅3^{n-1}\)
解析 证明:由\(S_{n+1}=3S_n+2\)得\(S_{n+1}+1=3\left(S_n+1\right)\),
即 \(\dfrac{s_{n+1}+1}{s_n+1}=3\),
又\(S_1+1=a_1+1=3\),
所以\(\{S_n+1\}\)是以\(3\)为首项,\(3\)为公比的等比数列,
解: 由(1) 可得\(S_n+1=3×3^{n-1}=3^n\), 即\(S_n=3^n-1\),
当\(n≥2\)时,\(a_n=S_n-S_{n-1}=3^n-1-\left(3^{n-1}-1\right)=2⋅3^{n-1}\).
又\(a_1=2\)满足\(a_n=2⋅3^{n-1}\).
所以\(a_n=2⋅3^{n-1}\). -
答案 \(a_n=\dfrac{n}{2^{n-1}}\)
解析 由 \(S_n=4-a_n-\dfrac{1}{2^{n-2}}\) 得 \(S_{n+1}=4-a_{n+1}-\dfrac{1}{2^{n-1}}\),
于是 \(S_{n+1}-S_n=\left(a_n-a_{n+1}\right)+\left(\dfrac{1}{2^{n-2}}-\dfrac{1}{2^{n-1}}\right)\),
所以 \(a_{n+1}=\left(a_n-a_{n+1}\right)+\dfrac{1}{2^{n-1}}\) , \(\therefore a_{n+1}=\dfrac{1}{2} a_n+\dfrac{1}{2^n}\) .
两边同乘以\(2^n\)得,\(2^n a_{n+1}=2^{n-1} a_n+1\),
则\(2^n a_{n+1}-2^{n-1} a_n=1\),
\(\therefore \{2^{n-1} a_n\}\)是公差为\(1\)的等差数列,
\(∵S_n=4-a_n-\dfrac{1}{2^{n-2}}\),
当\(n=1\)时,\(S_1=4-a_1-2\),\(\therefore a_1=1\),
\(\therefore 2^{n-1} a_n=1+1⋅{n-1}=n\), \(\therefore a_n=\dfrac{n}{2^{n-1}}\) . -
答案 (1) 略 ;(2) \(a_n=\dfrac{1}{2}×3^{n-1}\).
解析 (1)证明: 因为 \(a_{n+2}=2a_{n+1}+3a_n\),
所以\(a_{n+2}+a_{n+1}=3\left(a_{n+1}+a_n \right)\),
因为\(\{a_n\}\)中各项均为正数,所以\(a_{n+1}+a_n>0\),
所以 \(\dfrac{a_{n+2}+a_{n+1}}{a_{n+1}+a_n}=3\),
所以数列\(\{a_n+a_{n+1} \}\)是公比为\(3\)的等比数列,
(2) \(a_n+a_{n+1}=\left(a_1+a_2 \right) 3^{n-1}=2×3^{n-1}\),
因为\(a_{n+2}=2a_{n+1}+3a_n\),
所以\(a_{n+2}-3a_{n+1}=-\left(a_{n+1}-3a_n \right)\),\(a_2=3a_1\),
所以\(a_2-3a_1=0\),
所以\(a_{n+1}-3a_n=0\),故\(a_{n+1}=3a_n\),
所以\(4a_n=2×3^{n-1}\),即\(a_n=\dfrac{1}{2}×3^{n-1}\).
分层练习
【A组---基础题】
1.已知数列\(\{a_n\}\)的前\(n\)项和为\(S_n\),\(a_1=1\),满足下列条件
①\(∀n∈N^*\),\(a_n≠0\);\(\qquad \qquad\) ②点\(P_n \left(a_n,S_n\right)\)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\(n\)项和为\(S_n\).
2.已知数列\(\{a_n\}\)满足\(a_1=\dfrac{1}{2}\), \(a_{n+1}=a_n+\dfrac{1}{n^2+n}\),求\(a_n\).
3.已知\(a_1=1\),\(a_n=n\left(a_{n+1}-a_n \right) \left(n∈N^*\right)\),求数列\(\{a_n\}\)通项公式.
4.数列\(\{a_n\}\)中,若\(a_1=2\),且\(a_{n+1}=2a_n+2\),求数列\(\{a_n\}\)的通项公式.
5.若数列\(\{a_n\}\)中,\(a_1=1\),\(S_n\)是数列\(\{a_n\}\)的前\(n\)项之和,且 \(S_{n+1}=\dfrac{S_n}{3+4 S_n}(n \geq 1)\),求数列\(\{a_n\}\)的通项公式是\(a_n\).
6.记\(S_n\)是公差不为\(0\)的等差数列\(\{a_n\}\)的前n项和,已知\(a_3+3a_4=S_5\),\(a_1 a_5=S_4\),数列\(\{b_n \}\)满足\(b_n=3b_{n-1}+2^{n-1} \left(n≥2,n∈N^* \right)\),且\(b_1=a_1-1\),求\(\{a_n\}\), \(\{b_n \}\)的通项公式.
7.已知数列\(\{a_n\}\)满足\(a_1=3\),\(a_{n+1}=5a_n+3\cdot 2^n \left(n∈N^*\right)\),求数列\(\{a_n\}\)的通项公式.
8.正项数列\(\{a_n\}\)的前\(n\)项和为\(S_n\),已知\(a_1=a\),\(2S_n=a_n a_{n+1}\).
(1)若\(\{a_n\}\)是等差数列,求\(\{a_n\}\)的通项公式.
(2)是否存在实数\(a\),使得\(\{a_n\}\)是等比数列?若存在,求出\(a\)的值;若不存在,说明理由.
9.数列\(\{a_n\}\)前\(n\)项和\(S_n=4-a_n-\dfrac{1}{2^{n-2}}\),求通项公式\(a_n\).
10.数列 \(\{a_n\}\)\(n\)的前项和\(S_n\)满足\(a_1=3\),\(a_2=0\),\(a_n=6S_{n-2}\),\(n≥3\),\(n∈N^*\) .
(1)证明: 数列\(\{S_n+2S_{n-1} \}\)为等比数列;
(2)求\(\{a_n\}\)的通项公式.
参考答案
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答案 \(a_n=n\), \(S_n=\dfrac{n^2+n}{2}\)
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\(\left(a_n+a_{n-1}\right)\left(a_n-a_{n-1}-1\right)=0\),
又\(∀n∈N^*\),\(a_n≠0\),所以\(a_n+a_{n-1}=0\)或\(a_n-a_{n-1}-1=0\),
当\(a_n+a_{n-1}=0\)时,\(a_1=1\), \(\dfrac{a_n}{a_{n-1}}=-1\),
得 \(a_n=(-1)^{n-1}\), \(S_n=\dfrac{1-(-1)^n}{2}\);
当\(a_n-a_{n-1}-1=0\)时,\(a_1=1\),\(a_n-a_{n-1}=1\),
得\(a_n=n\), \(S_n=\dfrac{n^2+n}{2}\). -
答案 \(a_n=\dfrac{3}{2}-\dfrac{1}{n}\)
解析 由条件知: \(a_{n+1}-a_n=\dfrac{1}{n^2+n}=\dfrac{1}{n(n+1)}=\dfrac{1}{n}-\dfrac{1}{n+1}\)
\(\therefore a_2-a_1=1-\dfrac{1}{2}\),\(a_3-a_2=\dfrac{1}{2}-\dfrac{1}{3}\),\(a_4-a_3=\dfrac{1}{3}-\dfrac{1}{4}\),\(…\), \(a_{n-1}-a_{n-2}=\dfrac{1}{n-2}-\dfrac{1}{n-1}\),
\(a_n-a_{n-1}=\dfrac{1}{n-1}-\dfrac{1}{n}\),
把以上\(n-1\)个式子累加得到
\(a_n-a_1=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\cdots+\dfrac{1}{n-2}-\dfrac{1}{n-1}+\dfrac{1}{n-1}-\dfrac{1}{n}=1-\dfrac{1}{n}\),
\(\therefore a_n=a_1+1-\dfrac{1}{n}=\dfrac{3}{2}-\dfrac{1}{n}\). -
答案 \(a_n=n\)
解析 \(\because a_n=n\left(a_{n+1}-a_n \right)\), \(\therefore \dfrac{a_{n+1}}{a_n}=\dfrac{n+1}{n} \text {, }\),
又有 \(a_n=a_1 \cdot \dfrac{a_2}{a_1} \cdot \dfrac{a_3}{a_2} \ldots \dfrac{a_n}{a_{n-1}}=1 \times \dfrac{2}{1} \times \dfrac{3}{2} \times \ldots \times \dfrac{n}{n-1}=n\),
当\(n=1\)\(a_1=1\)时,满足\(a_n=n\),\(\therefore a_n=n\). -
答案 \(a_n=2^{n+1}-2\)
解析 \(\because a_{n+1}=2a_n+2\),\(\therefore a_{n+1}+2=2a_n+2+2=2\left(a_n+2\right)\),
\(\therefore \dfrac{a_{n+1}+2}{a_n+2}=2\)为常数.
\(\therefore \{a_n+2\}\)为等比数列,首项\(a_1+2=4\),公比\(q=2\),
\(\therefore a_n+2=4×2^{n-1}=2^{n+1}\);
\(\therefore a_n=2^{n+1}-2\);
\(\therefore\) 数列\(\{a_n\}\)的通项公式为\(a_n=2^{n+1}-2\). -
答案 \(a_n=\left\{\begin{array}{c} 1, n=1 \\ \dfrac{1}{3^n-2}-\dfrac{1}{3^{n-1}-2}, n \geq 2 \end{array}\right.\)
解析 递推式 \(S_{n+1}=\dfrac{S_n}{3+4 S_n}\)可变形为 \(\dfrac{1}{s_{n+1}}=3 \cdot \dfrac{1}{s_n}+4\),
则 \(\dfrac{1}{s_{n+1}}+2=3 \cdot\left(\dfrac{1}{s_n}+2\right)\),
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\(n≥2\), \(a_n=S_n-S_{n-1}=\dfrac{1}{3^n-2}-\dfrac{1}{3^{n-1}-2},\),
而\(a_1=1\)不满足 \(a_n=\dfrac{1}{3^n-2}-\dfrac{1}{3^{n-1}-2}\),
\(\therefore a_n=\left\{\begin{array}{c} 1, n=1 \\ \dfrac{1}{3^n-2}-\dfrac{1}{3^{n-1}-2}, n \geq 2 \end{array}\right.\) . -
答案\(a_n=2n\),\(b_n=3^n-2^n\).
解析 设等差数列\(\{a_n\}\)的公差为\(d(d≠0)\),因为\(a_3+3a_4=S_5\),\(a_1 a_5=S_4\),
则 \(\left\{\begin{array}{l} a_1+2 d+3 a_1+9 d=5 a_1+10 d \\ a_1\left(a_1+4 d\right)=4 a_1+6 d \end{array}\right.\),
解得 \(\left\{\begin{array}{l} a_1=2 \\ d=2 \end{array}\right.\)或 \(\left\{\begin{array}{l} a_1=0 \\ d=0 \end{array}\right.\)(舍),
所以\(a_n=2n\),
因为\(b_n=3b_{n-1}+2^{n-1} \left(n≥2,n∈N^* \right)\),
所以 \(\dfrac{b_n}{2^n}=\dfrac{3}{2} \cdot \dfrac{b_{n-1}}{2^{n-1}}+\dfrac{1}{2}\),即 \(\dfrac{b_n}{2^n}+1=\dfrac{3}{2}\left(\dfrac{b_{n-1}}{2^{n-1}}+1\right)\),
所以 \(\dfrac{\dfrac{b_n}{2^n}+1}{\dfrac{b_{n-1}}{2^{n-1}}+1}=\dfrac{3}{2}\),
因为\(b_1=a_1-1=1\),所以\(\dfrac{b_1}{2}+1=\dfrac{3}{2}\).
所以数列 \(\left\{\dfrac{b_n}{2^n}+1\right\}\)是以\(\dfrac{3}{2}\)为首项,\(\dfrac{3}{2}\)为公比的等比数列.
所以 \(\dfrac{b_n}{2^n}+1=\left(\dfrac{3}{2}\right)^n\),所以\(b_n=3^n-2^n\). -
答案 \(a_n=5^n-2^n\)
解析 数列\(\{a_n\}\)满足\(a_1=3\),\(a_{n+1}=5a_n+3\cdot 2^n \left(n∈N^*\right)\).
整理得\(a_{n+1}+2^{n+1}=5\left(a_n+2^n\right)\),
令\(b_n=a_n+2^n\),则\(b_{n+1}=5b_n\),
当\(n=1\)时,\(b_1=5\),
所以数列\(\{b_n \}\)是以\(5\)为首项、\(5\)为公比的等比数列.
\(\therefore b_n=5^n\),则\(a_n=5^n-2^n\). -
答案 (1)\(a_n=n\);(2) 不存在实数\(a\),使得\(\{a_n\}\)是等比数列.
解析 (1)当\(n=1\)时,\(2S_1=2a_1=a_1 a_2\),又\(a_1>0\),则\(a_2=2\),
当\(n≥2\)时,\(2a_n=2S_n-2S_{n-1}=a_n a_{n+1}-a_{n-1} a_n\),即\(a_{n+1}-a_{n-1}=2\),
因为\(\{a_n\}\)是等差数列,设\(\{a_n\}\)的公差为\(d\),
所以\(a_{n+1}-a_{n-1}=2d=2\),解得\(d=1\),
则\(a_1=a_2-d=2-1=1\),故\(\{a_n\}\)的通项公式为\(a_n=n\);
(2)假设存在实数\(a\),使得\(\{a_n\}\)是等比数列,
由 (1)可知,\(a_3=a_1+2=2+a\),\(a_4=a_2+2=4\),
因为\(\{a_n\}\)是等比数列,所以\(a_2 a_3=a_1 a_4\),即\(2 \left(2+a\right)=4a\),解得\(a=2\),
此时 \(\dfrac{a_2}{a_1}=\dfrac{2}{2}=1\), \(\dfrac{a_3}{a_2}=\dfrac{4}{2}=2\)不符合题意,则假设错误.
故不存在实数\(a\),使得\(\{a_n\}\)是等比数列. -
答案 \(a_n=\dfrac{n}{2^{n-1}}\)
解析 由\(S_n=4-a_n-\dfrac{1}{2^{n-2}}\) 得: \(S_{n+1}=4-a_{n+1}-\dfrac{1}{2^{n-1}}\)
于是\(S_{n+1}-S_n=\left(a_n-a_{n+1}\right)+\left(\dfrac{1}{2^{n-2}}-\dfrac{1}{2^{n-1}}\right)\)
所以\(a_{n+1}=\left(a_n-a_{n+1}\right)+\dfrac{1}{2^{n-1}}\) , \(\therefore a_{n+1}=\dfrac{1}{2} a_n+\dfrac{1}{2^n}\).
两边同乘以\(2^n\)得,\(2^n a_{n+1}=2^{n-1} a_n+1\),则\(2^n a_{n+1}-2^{n-1} a_n=1\),
\(\therefore \{2^{n-1} a_n\}\)是公差为\(1\)的等差数列,
\(\because S_n=4-a_n-\dfrac{1}{2^{n-2}}\),
当\(n=1\)时,\(S_1=4-a_1-2\),\(\therefore a_1=1\),
\(\therefore 2^{n-1} a_n=1+1⋅(n-1)=n\), \(\therefore a_n=\dfrac{n}{2^{n-1}}\). -
答案 (1)略;(2) \(a_n=\dfrac{2}{5} \cdot 3^n+\dfrac{9}{5} \cdot(-2)^{n-1}\)
解析 由数列\(\{a_n\}\)的前\(n\)项和\(S_n\),满足\(a_n=6S_{n-2}\),\(n≥3\),\(n∈N^*\)
因为\(a_n=S_n-S_{n-1}\),可得 \(S_n-S_{n-1}=6S_{n-2}\),
即\(S_n+2S_{n-1}=3S_{n-1}+6S_{n-2}=3\left(S_{n-1}+2S_{n-2} \right)\),
即 \(\dfrac{S_n+2 S_{n-1}}{S_{n-1}+2 S_{n-2}}=3\left(n \geq 3, n \in N^*\right)\),
又由\(a_1=3\),\(a_2=0\),可得\(2S_1+S_2=3a_1+a_2=9\),
所以数列\(\{S_n+2S_{n-1} \}\)表示首项为\(9\),公比为\(3\)的等比数列.
解: 由(1)可得\(S_n+2S_{n-1}=9⋅3^{n-2}=3^n\),
当\(n≥3\)时,可得\(S_{n-1}+2S_{n-2}=3^{n-1}\),
两式相减得\(\left(S_n-S_{n-1} \right)+2\left(S_{n-1}-S_{n-2} \right)=3^n-3^{n-1}=2⋅3^{n-1}\),
即\(a_n+2a_{n-1}=2⋅3^{n-1}\),
即 \(\dfrac{a_n}{3^n}+\dfrac{2}{3} \cdot \dfrac{a_{n-1}}{3^{n-1}}=\dfrac{2}{3}\),即 \(\dfrac{a_n}{3^n}=-\dfrac{2}{3} \cdot \dfrac{a_{n-1}}{3^{n-1}}+\dfrac{2}{3}\),
设 \(b_n=\dfrac{a_n}{3^n}\),
可得 \(b_n=-\dfrac{2}{3} \cdot b_{n-1}+\dfrac{2}{3}\),即 \(b_n-\dfrac{2}{5}=-\dfrac{2}{3}\left(b_{n-1}-\dfrac{2}{5}\right)\),
又由\(a_1=3\),\(a_2=0\),
可得 \(b_1-\dfrac{2}{5}=\dfrac{a_1}{3}-\dfrac{2}{5}=\dfrac{3}{5}\), \(b_2-\dfrac{2}{5}=\dfrac{a_2}{3^2}-\dfrac{2}{5}=-\dfrac{2}{5}\),
因为\(a_n=6S_{n-2},n≥3,n∈N^*\),
所以\(a_3=6S_1=18\),可得 \(b_3-\dfrac{2}{5}=\dfrac{a_3}{3^3}-\dfrac{2}{5}=\dfrac{4}{15}\),
所以数列\(\{b_n-\dfrac{2}{5}\}\)表示首项为\(\dfrac{3}{5}\),公比为\(-\dfrac{2}{3}\)的等比数列,
所以\(b_n-\dfrac{2}{5}=\dfrac{3}{5} \cdot\left(-\dfrac{2}{3}\right)^{n-1}\),即\(\dfrac{a_n}{3^n}-\dfrac{2}{5}=\dfrac{3}{5} \cdot\left(-\dfrac{2}{3}\right)^{n-1}\),
即\(a_n=\dfrac{2}{5} \cdot 3^n+\dfrac{9}{5} \cdot(-2)^{n-1}\).
【B组---提高题 】
1.已知数列\(\{a_n\}\)的前\(n\)项和\(S_n\),满足\(a_2=-4\),\(2S_n=n\left(a_n-7\right)\).求\(a_1\)和数列\(\{a_n\}\)的通项公式.
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\(a_n\).
参考答案
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答案 \(a_n=3n-10\)
解析 在\(2S_n=n\left(a_n-7\right)\)中,
当\(n=1\)时,\(2S_1=a_1-7⇒2a_1=a_1-7⇒a_1=-7\).
由\(2S_n=n\left(a_n-7\right)\)得,\(2S_{n+1}=(n+1)\left(a_{n+1}-7\right)\),
两式相减得,\(2a_{n+1}=(n+1) a_{n+1}-na_n-7\),
所以\((n-1) a_{n+1}-na_n=7\),
当\(n≥2\)时,有 \(\dfrac{a_{n+1}}{n}-\dfrac{a_n}{n-1}=\dfrac{7}{n(n-1)}=7\left(\dfrac{1}{n-1}-\dfrac{1}{n}\right)\)
所以 \(\dfrac{a_{n+1}}{n}-\dfrac{a_2}{1}=\left(\dfrac{a_{n+1}}{n}-\dfrac{a_n}{n-1}\right)+\left(\dfrac{a_n}{n-1}-\dfrac{a_{n-1}}{n-2}\right)+\cdots \cdots+\left(\dfrac{a_3}{2}-\dfrac{a_2}{1}\right)\)
\(=7\left(\dfrac{1}{n-1}-\dfrac{1}{n}\right)+7\left(\dfrac{1}{n-2}-\dfrac{1}{n-1}\right)+\cdots \ldots+7\left(\dfrac{1}{1}-\dfrac{1}{2}\right)\)
\(=7\left(1-\dfrac{1}{n}\right)=\dfrac{7(n-1)}{n}\),
所以\(\dfrac{a_{n+1}}{n}=\dfrac{a_2}{1}+\dfrac{7(n-1)}{n}=-4+\dfrac{7(n-1)}{n}=\dfrac{3 n-7}{n}\),
所以 \(a_{n+1}=3n-7\left(n≥2\right)\),
故\(a_n=3n-10\left(n≥3\right)\),
又\(a_1=-7\),\(a_2=-4\)也都符合上式,
所以\(a_n=3n-10\left(n∈N^*\right)\). -
答案 \(a_n=\dfrac{1}{n}\)
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\([(n+1) a_{n+1}-na_n]\left(a_{n+1}+a_n\right)=0\)
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 \(a_n=\dfrac{n-1}{n} a_{n-1}=\dfrac{n-1}{n} \cdot \dfrac{n-2}{n-1} a_{n-2}=\cdots=\dfrac{n-1}{n} \cdot \dfrac{n-2}{n-1} \cdots \dfrac{2}{3} \cdot \dfrac{1}{2} a_1\)
\(=\dfrac{1}{n} a_1=\dfrac{1}{n}(n≥2)\),
\(a_1=1\)也满足 \(a_n=\dfrac{1}{n}\),
\(\therefore a_n=\dfrac{1}{n}, \quad n \in N^*\).
【C组---拓展题 】
1.已知数列\(\{a_n\}\)的各项都是正数,且满足\(a_1=1\),\(a_{n+1}=\dfrac{1}{2} a_n \left(4-a_n \right)\) ,\(n∈N\).求数列\(\{a_n\}\)的通项公式\(a_n\).
2.已知数列\(\{a_n\}\)的前\(n\)项和为\(S_n\),\(a_n>0\)且 \(S_n=\dfrac{1}{2}\left(a_n+\dfrac{n}{a_n}\right)\), 求数列\(\{a_n\}\)通项公式.
参考答案
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答案 \(a_n=2-\left(\dfrac{1}{2}\right)^{2^{n-1}-1}\)
解析 \(a_{n+1}=\dfrac{1}{2} a_n\left(4-a_n\right)=\dfrac{1}{2}\left[-\left(a_n-2\right)^2+4\right]\),
所以\(a_{n+1}-2=-\dfrac{1}{2}\left(a_n-2\right)^2\)
令\(b_n=a_n-2\),
则 \(b_n=-\dfrac{1}{2}\left(b_{n-1}\right)^2=-\dfrac{1}{2}\left[-\dfrac{1}{2}\left(b_{n-2}\right)^2\right]^2=-\left(\dfrac{1}{2}\right)^{1+2}\left(b_{n-2}\right)^{2^2}=-\left(\dfrac{1}{2}\right)^{1+2+4}\left(b_{n-3}\right)^{2^3}\) \(=\cdots=-\left(\dfrac{1}{2}\right)^{1+2+4+\cdots+2^{n-2}}\left(b_1\right)^{2^{n-1}}\),
又\(b_1=-1\),
所以\(b_n=-\left(\dfrac{1}{2}\right)^{2^{n-1}-1}\),即 \(a_n=2+b_n=2-\left(\dfrac{1}{2}\right)^{2^{n-1}-1}\). -
答案 \(a_n=\sqrt{\dfrac{n(n+1)}{2}}-\sqrt{\dfrac{n(n-1)}{2}}\)
解析 \(\because S_n=\dfrac{1}{2}\left(a_n+\dfrac{n}{a_n}\right)\),
当\(n=1\)时,有 \(S_1=\dfrac{1}{2}\left(a_1+\dfrac{1}{a_1}\right)\),解得\(a_1=1\),
当\(n≥2\)时, \(S_n=\dfrac{1}{2}\left(S_n-S_{n-1}+\dfrac{n}{S_n-S_{n-1}}\right)\),
化简为\(S_n^2-S_{n-1}^2=n\),
则 \(S_n^2=\left(S_n^2-S_{n-1}^2\right)+\left(S_{n-1}^2-S_{n-2}^2\right)+\cdots+\left(S_2^2-S_1^2\right)+S_1^2\)
\(=n+(n-1)+\cdots+2+1=\dfrac{n(n+1)}{2}(n \geq 2)\),
\(\because a_n>0\), \(\therefore S_n=\sqrt{\dfrac{n(n+1)}{2}}(n \geq 2)\),
而\(S_1=a_1=1\)也满足 \(S_n=\sqrt{\dfrac{n(n+1)}{2}}(n \geq 2)\),
故\(S_n=\sqrt{\dfrac{n(n+1)}{2}}\left(n \in N^*\right)\),
则\(a_n=\sqrt{\dfrac{n(n+1)}{2}}-\sqrt{\dfrac{n(n-1)}{2}}\).