专题 求数列的前n项和

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知识剖析

求数列的前项和是数列中常考的一大专题，其方法有公式法、倒序相加(乘)法、分组求和法与裂项相消法等，在掌握这些方法的时候要注意方法的适用范围，其中的计算量有些大，技巧性也较强，需要多加以理解与总结.

经典例题

【方法一】公式法

若已知数列是等差或等比数列，求其前\(n\)项和可直接使用对应的公式；若求和的式子对应某些公式，也可以直接使用.常见如下

\((1)\)等差数列求和公式\(S_{n}=\dfrac{n\left(a_{1}+a_{n}\right)}{2}=n a_{1}+\dfrac{n(n-1)}{2} d\)

\((2)\)等比数列求和公式\(S_{n}=\left\{\begin{array}{l}
n a_{1}, q=1 \\
\dfrac{a_{1}\left(1-q^{n}\right)}{1-q}, q \neq 1
\end{array}\right.\)

\((3)\)\(1^{2}+2^{2}+3^{2}+\cdots+n^{2}=\dfrac{n(n+1)(2 n+1)}{6}\)

\((4)\)\(1^{3}+2^{3}+3^{3}+\cdots+n^{3}=\left[\dfrac{n(n+1)}{2}\right]^{2}\)

【典题1】求和式\(3+6+12+\cdots+3 \cdot 2^{n-2}\)，先思考它是几项之和再求和.

【解析】和式\(3+6+12+\cdots+3 \cdot 2^{n-2}\)相当于数列\(3\)、\(6\)、\(12\)、…、\(3 \cdot 2^{n-2}\)的和，

显然它是首项\(a_1=3\)，公比\(q=2\)的等比数列，

设前\(n\)项和为\(S_n\)，

故\(a_{n}=a_{1} \cdot q^{n-1}=3 \cdot 2^{n-1}\)，

而和式最后一项是\(3 \cdot 2^{n-2}=a_{n-1}\)，是第\(n-1\)项，

故和式\(3+6+12+\cdots+3 \cdot 2^{n-2}\)只有\(n-1\)项而已，

则\(3+6+12+\cdots+3 \cdot 2^{n-2}\)

\({\color{Red}{ (切勿想当然和式等于S_n)}}\)

\(=S_{n-1}=\dfrac{a_{1}\left(1-q^{n-1}\right)}{1-q}=\dfrac{3\left(1-2^{n-1}\right)}{1-2}=3\left(2^{n-1}-1\right)\).

【点拨】求和式时特别要注意确定项数，以第一个数为首项，判断最后一项为第几项(第\(n\)项、第\(n-1\)项？)便可.

【典题2】已知等比数列\(\left\{a_{n}\right\}\)前\(n\)项和为\(S_n\)，且\(S_{n}=a_{n+1}-\dfrac{1}{32}\left(n \in \boldsymbol{N}^{*}\right)\)．

(1)求数列\(\left\{a_{n}\right\}\)的通项公式；

(2)若\(b_{n}=\log _{2} a_{n}\)，求数列\(\left\{\left|b_{n}\right|\right\}\)的前\(n\)项和\(T_n\)．

【解析】(1)由于\(S_{n}=a_{n+1}-\dfrac{1}{32}\)①，

当\(n=1\)时，\(S_{1}=a_{2}-\dfrac{1}{32} \Rightarrow a_{2}=a_{1}+\dfrac{1}{32}\)，

当\(n≥2\)时，\(S_{n-1}=a_{n}-\dfrac{1}{32}\)②，

①－②得\(a_{n}=a_{n+1}-a_{n}\)，即\(a_{n+1}=2 a_{n}(n \geq 2)\)

\(∵\)数列\(\left\{a_{n}\right\}\)为等比数列，

\(∴a_2=2a_1\)，又\(a_{2}=a_{1}+\dfrac{1}{32}\)，解得\(a_{1}=\dfrac{1}{32}\)．

故数列\(\left\{a_{n}\right\}\)是以\(\dfrac{1}{32}\)为首项，\(2\)为公比的等比数列，

所以\(a_{n}=2^{n-6}\)．

(2)\(b_{n}=\log _{2} a_{n}=n-6\)，

所以\(\left|b_{n}\right|=\left\{\begin{array}{l}
6-n, n<6 \\
n-6, n \geq 6
\end{array}\right.\)，

\({\color{Red}{(遇到绝对值，则可利用|x|=\left\{\begin{array}{c}
x, x \geq 0 \\
-x, x<0
\end{array}\right.去掉绝对值，则求前n项和T_n时要注意分类讨论)
}}\)

当\(n<6\)时，

\(T_{n}=-b_{1}-b_{2}-\cdots-b_{n}=-\left(b_{1}+b_{2}+\cdots+b_{n}\right)\)

\(=-\left[-5 n+\dfrac{n(n-1)}{2}\right]=-\dfrac{n^{2}-11 n}{2}=\dfrac{11 n-n^{2}}{2}\)

\({\color{Red}{(b_n=n-6是等差数列，可由前n项和公式S_{n}=n a_{1}+\dfrac{n(n-1)}{2} d得b_{1}+\cdots+b_{n}=\dfrac{n^{2}-11 n}{2})
}}\)

当\(n≥6\)时，

\(\begin{aligned}
T_{n} &=-b_{1}-\cdots-b_{5}+b_{6}+\cdots+b_{n} \\
&=\left(b_{1}+b_{2}+\cdots+b_{n}\right)-2\left(b_{1}+\cdots+b_{5}\right) \\
&=\dfrac{n^{2}-11 n}{2}-2 \times \dfrac{5^{2}-11 \times 5}{2}=\dfrac{n^{2}-11 n}{2}+30
\end{aligned}\)

\(\therefore T_{n}=\left\{\begin{array}{l}
\dfrac{11 n-n^{2}}{2}, n<6 \\
\dfrac{n^{2}-11 n}{2}+30, n \geq 6
\end{array}\right.\)．

【点拨】当确保数列为等差数列或等比数列，便可直接使用对应的前\(n\)项和公式，这需要明确等差数列通项公式形如\(a_n=kn+b\)，等比数列通项公式形如\(a_n=A\cdot B^n\).

巩固练习

1(★★)求和式\(1+4+7+⋯+(3n+1)\).

 

 

2(★★)已知\(\left\{a_{n}\right\}\)是等差数列，公差\(d≠0\)，\(a_1=1\)，且\(a_1\),\(a_3\),\(a_9\)成等比数列，求数列\(\left\{2^{a_{n}}\right\}\)的前\(n\)项和\(S_n\)．

3(★★)已知等差数列\(\left\{a_{n}\right\}\)前三项的和为\(-3\)，前三项的积为\(15\)，

(1)求等差数列\(\left\{a_{n}\right\}\)的通项公式；

(2)若公差\(d>0\)，求数列\(\left\{\left|a_{n}\right|\right\}\)的前\(n\)项和\(T_n\)．

4(★★★)设\(\left\{a_{n}\right\}\)是公比大于\(1\)的等比数列，\(S_n\)为数列\(\left\{a_{n}\right\}\)的前\(n\)项和．已知\(S_3=7\)，且\(a_1+3\)，\(3a_2\)，\(a_3+4\)构成等差数列．

(1)求数列\(\left\{a_{n}\right\}\)的等差数列．

(2)令\(b_{n}=\ln a_{3 n+1}\)，求数列\(\left\{b_{n}\right\}\)的前\(n\)项和\(T_n\)．

参考答案

1.\(\dfrac{3 n^{2}+5 n+2}{2}\)

2.\(S_{n}=2^{n+1}-2\)

3.\((1) a_n=4n－9\)或\(a_n=7－4n\)，\((2) T_{n}= \begin{cases}5, & n=1 \\ 2 n^{2}-7 n+12, & n \geq 2\end{cases}\)

4.\((1) a_{n}=2^{n-1}\)，\((2) T_{n}=\dfrac{3 \ln 2}{2} n(n+1)\)

【方法二】 倒序相加(乘)法

1 对于某个数列\(\left\{a_{n}\right\}\)，若满足\(a_{1}+a_{n}=a_{2}+a_{n-1}\)\(=\cdots=a_{k}+a_{n-k+1}\)，则求前\(n\)项和\(S_n\)可使用倒序相加法.

具体解法：设\(S_{n}=a_{1}+a_{2}+\cdots+a_{n-1}+a_{n}\)①

把①反序可得\(S_{n}=a_{n}+a_{n-1}+\cdots+a_{2}+a_{1}\)②

由①+②得\(2 S_{n}=\left(a_{1}+a_{n}\right)+\left(a_{2}+a_{n-1}\right)+\cdots+\left(a_{n-1}+a_{2}\right)+\left(a_{n}+a_{1}\right)\)\(\Rightarrow S_{n}=\dfrac{\left(a_{1}+a_{n}\right) n}{2}\).

2 对于某个数列\(\left\{a_{n}\right\}\)，若满足\(a_{1} a_{n}=a_{2} a_{n-1}=\cdots=a_{k} a_{n-k+1}\)，则求前\(n\)项积\(T_n\)可使用倒序相乘法.具体解法类同倒序相加法.

【典题1】设\(f(x)=\dfrac{1}{4^{x}+2}\)，利用课本中推导等差数列前\(n\)项和的公式的方法，可求得\(f(-3)+f(-2)+⋯+f(0)+⋯+f(3)+f(4)\)的值为\(\underline{\quad \quad}\)．

【解析】设\(a+b＝1\)，

则\(f(a)+f(b)=\dfrac{1}{4^{a}+2}+\dfrac{1}{4^{b}+2}\)\(=\dfrac{4^{b}}{\left(4^{a}+2\right) 4^{b}}+\dfrac{1}{4^{b}+2}\)\(=\dfrac{4^{b}}{4+2 \cdot 4^{b}}+\dfrac{1}{4^{b}+2}\)\(=\dfrac{4^{b}+2}{2\left(4^{b}+2\right)}=\dfrac{1}{2}\)．

所以\(f(-3)+f(4)=\dfrac{1}{2}\)，\(f(-2)+f(3)=\dfrac{1}{2}\)，\(f(-1)+f(2)=\dfrac{1}{2}\)，\(f(0)+f(1)=\dfrac{1}{2}\)，

\(f(-3)+f(-2)+\cdots+f(0)+\cdots+f(3)+f(4)\)\(=4 \times \dfrac{1}{2}=2\)．

【点拨】课本中推导等差数列前\(n\)项和的公式的方法就是倒序相加法.

【典题2】求\(\sin ^{2} 1^{\circ}+\sin ^{2} 2^{\circ}+\sin ^{2} 3^{\circ}+\cdots+\sin ^{2} 88^{\circ}+\sin ^{2} 89^{\circ}\)的值

【解析】设\(\sin ^{2} 1^{\circ}+\sin ^{2} 2^{\circ}+\sin ^{2} 3^{\circ}+\cdots+\sin ^{2} 88^{\circ}+\sin ^{2} 89^{\circ}\)…………. ①

将①式右边反序得

\(S=\sin ^{2} 89^{\circ}+\sin ^{2} 88^{\circ}+\cdots+\sin ^{2} 3^{\circ}+\sin ^{2} 2^{\circ}+\sin ^{2} 1^{\circ}\)…………..②

①+②得

\(\begin{aligned}
2 S &=\left(\sin ^{2} 1^{\circ}+\sin ^{2} 89^{\circ}\right)+\left(\sin ^{2} 2^{\circ}+\sin ^{2} 88^{\circ}\right)+\cdots+\left(\sin ^{2} 89^{\circ}+\sin ^{2} 1^{\circ}\right) \\
&=\left(\sin ^{2} 1^{\circ}+\cos ^{2} 1^{\circ}\right)+\left(\sin ^{2} 2^{\circ}+\cos ^{2} 2^{\circ}\right)+\cdots+\left(\sin ^{2} 89^{\circ}+\cos ^{2} 89^{\circ}\right) \\
&=89
\end{aligned}\)

\(∴S=44.5\).

【点拨】对于某个数列\(\left\{a_{n}\right\}\)，若满足\(a_{1}+a_{n}\)\(=a_{2}+a_{n-1}=\cdots=a_{k}+a_{n-k+1}\)，则可使用倒序相加法.

【典题3】设函数\(f(x)=\dfrac{2^{x}}{2^{x}+\sqrt{2}}\)的图象上两点\(P_1 (x_1 ,y_1)\)、\(P_2 (x_2 ,y_2)\)，若\(\overrightarrow{O P}=\dfrac{1}{2}\left(\overrightarrow{O P_{1}}+\overrightarrow{O P_{2}}\right)\)，且点\(P\)的横坐标为\(\dfrac{1}{2}\)．

(1)求证：\(P\)点的纵坐标为定值，并求出这个定值；

(2)求\(S_{n}=f\left(\dfrac{1}{n}\right)+f\left(\dfrac{2}{n}\right)+\cdots+f\left(\dfrac{n-1}{n}\right)+f\left(\dfrac{n}{n}\right)\).

【解析】(1)证：\(∵\overrightarrow{O P}=\dfrac{1}{2}\left(\overrightarrow{O P_{1}}+\overrightarrow{O P_{2}}\right)\)，

\(∴P\)是\(P_1 P_2\)的中点\(⇒x_1+x_2=1\)

\(\therefore y_{1}+y_{2}=f\left(x_{1}\right)+f\left(x_{2}\right)\)\(=\dfrac{2^{x_{1}}}{2^{x_{1}}+\sqrt{2}}+\dfrac{2^{x_{2}}}{2^{x_{2}}+\sqrt{2}}\)\(=\dfrac{2^{x_{1}}}{2^{x_{1}}+\sqrt{2}}+\dfrac{2^{1-x_{1}}}{2^{1-x_{1}}+\sqrt{2}}\)\(=\dfrac{2^{x_{1}}}{2^{x_{1}}+\sqrt{2}}+\dfrac{2}{\sqrt{2} \cdot 2^{x_{1}+2}}=1\)．

\(\therefore y_{p}=\dfrac{1}{2}\left(y_{1}+y_{2}\right)=\dfrac{1}{2}\)．

(2)解：由(1)知\(x_1+x_2=1\)，\(f (x_1)+f (x_2)=y_1+y_2=1\)，\(f(1)=2-\sqrt{2}\)，

\({\color{Red}{(即横坐标之和为1，则对应的坐标之和为1，则有f\left(\dfrac{1}{n}\right)+f\left(\dfrac{n-1}{n}\right)=f\left(\dfrac{k}{n}\right)+f\left(\dfrac{n-k}{n}\right)=1，想到倒序相加法)}}\)

由\(S_{n}=f\left(\dfrac{1}{n}\right)+f\left(\dfrac{2}{n}\right)+\cdots+f\left(\dfrac{n-1}{n}\right)+f\left(\dfrac{n}{n}\right)\)

得\(S_{n}=f\left(\dfrac{n}{n}\right)+f\left(\dfrac{n-1}{n}\right)+\cdots+f\left(\dfrac{2}{n}\right)+f\left(\dfrac{1}{n}\right)\)

两式相加得

\(\begin{aligned}
2 S_{n}=& f(1)+\left[f\left(\dfrac{1}{n}\right)+f\left(\dfrac{n-1}{n}\right)\right]+\left[f\left(\dfrac{2}{n}\right)+f\left(\dfrac{n-2}{n}\right)\right]+\cdots +\left[f\left(\dfrac{n-1}{n}\right)+f\left(\dfrac{1}{n}\right)\right]+f(1) \\
=& 2 f(1)+n-1 \\
=& n+3-2 \sqrt{2}
\end{aligned}\)

\(\therefore S_{n}=\dfrac{n+3-2 \sqrt{2}}{2}\)．

巩固练习

1(★★)设等差数列\(\left\{a_{n}\right\}\)，公差为\(d\)，求证：\(\left\{a_{n}\right\}\)的前\(n\)项和\(S_{n}=\dfrac{\left(a_{1}+a_{n}\right) n}{2}\).

2(★★)设\(f(x)＝(x-1)^3+1\)，求\(f(-4)+⋯\)\(+f(0)+⋯+f(5)+f(6)\)的值为\(\underline{\quad \quad}\)

3(★★)设函数\(f(x)=\dfrac{x^{2}}{1+x^{2}}\)，求\(f(1)+f(2)\)\(+f\left(\dfrac{1}{2}\right)+f(3)+f\left(\dfrac{1}{3}\right)+f(4)+f\left(\dfrac{1}{4}\right)\)的值\(\underline{\quad \quad}\)．

参考答案

1.提示：倒序相加法

2.\(11\)

3.\(\dfrac{7}{2}\)

【方法三】 分组求和法

1 若数列\(\{c_n\}\)中通项公式\(c_n=a_n+b_n\)，可分成两个数列\(\left\{a_{n}\right\}\)，\(\{b_n\}\)之和，则数列\(\{c_n\}\)的前\(n\)项和等于两个数列\(\left\{a_{n}\right\}\)，\(\{b_n\}\)的前\(n\)项和的和.

2 常见的是\(c_n=\)等差+等比形式，分组求和法的解题套路如下



3 等比数列的通项公式形如\(a_n=kn+b\)，等差数列的通项公式形如\(a_n=A\cdot B^n\).

【典题1】求数列\(\left\{3^{n}+2 n-1\right\}\)的前\(n\)项和为\(S_n\).

【解析】设\(a_n=3^n+2n-1\)，

\({\color{Red}{(数列\left\{3^{n}\right\}是等比数列， \{2 n-1\}是等差数列)}}\)

则\(S_n=a_1+a_2+a_3+⋯+a_n\)

\(=(3^1+1)+(3^2+3)+(3^3+5)+⋯(3^n+2n-1)\)

\({\color{Red}{(把等比项和等差项分别放在一组) }}\)

\(=(3^1+3^2+3^3+⋯+3^n )+(1+3+5+⋯+2n-1)\)

\({\color{Red}{(确定好首项和公差、公比) }}\)

\(=\dfrac{3\left(1-3^{n}\right)}{1-3}+\dfrac{(1+2 n-1) n}{2}\)

\(=\dfrac{3^{n+1}}{2}+n^{2}-\dfrac{3}{2}\).

【典题2】已知等差数列\(\left\{a_{n}\right\}\)的前\(n\)项和为\(S_n\)，且\(a_5=5a_1\)，\(S_3-a_2=8\)．

(1)求数列\(\left\{a_{n}\right\}\)的通项公式；

(2)若数列\(\{b_n\}\)满足\((n×2^n+S_n)b_n=a_n\)，求数列\(\left\{\dfrac{1}{b_{n}}\right\}\)的前\(n\)项和\(T_n\)．

【解析】(1)等差数列\(\left\{a_{n}\right\}\)的前\(n\)项和为\(S_n\)，

设公差为\(d\)，且\(a_5=5a_1\),\(S_3－a_2=8\)．

\(\therefore\left\{\begin{array}{l}
a_{1}+4 d=5 a_{1} \\
2 a_{1}+2 d=8
\end{array}\right.\)，解得\(\left\{\begin{array}{c}
a_{1}=2 \\
d=2
\end{array}\right.\)，

故\(a_n=2n\)；

(2)由于\(a_n=2n\)，

\(\therefore S_{n}=\dfrac{(2+2 n) n}{2}=n^{2}+n\)，

又\(∵\)数列\(\{b_n\}\)满足\(\left(n 2^{n}+S_{n}\right) b_{n}=a_{n}\)，

\(\therefore \dfrac{1}{b_{n}}=\dfrac{2^{n}+n+1}{2}\)，

则\(T_{n}=\dfrac{1}{b_{1}}+\dfrac{1}{b_{2}}+\cdots+\dfrac{1}{b_{n}}\)\(=\dfrac{1}{2}\left[\left(2^{1}+2^{2}+\ldots+2^{n}\right)+\left(\dfrac{n(n+1)}{2}+n\right)\right]\)\(=2^{n}+\dfrac{n^{2}}{4}+\dfrac{3 n}{4}-1\)．

【典题3】设数列\(\left\{a_{n}\right\}\)满足\(a_1=1\)，\(\dfrac{a_{n+1}}{a_{n}}=2^{n}\)\((n∈N^*)\)．

(1)求数列\(\left\{a_{n}\right\}\)的通项公式；

(2)设\(b_{n}=\log _{2} a_{n}\)，求数列\(b_2+b_3+⋯+b_{100}\)的值．

【解析】(1)数列\(\left\{a_{n}\right\}\)满足\(a_1=1\)，\(\dfrac{a_{n+1}}{a_{n}}=2^{n}\)\((n∈N^*)\)．

\(\therefore a_{n}=\left(\dfrac{a_{n}}{a_{n-1}} \cdot \dfrac{a_{n-1}}{a_{n-2}} \cdots \dfrac{a_{2}}{a_{1}}\right) \cdot a_{1}(n \geq 2)\)，

\(\therefore a_{n}=\left(2^{n-1} \cdot 2^{n-2} \cdots 2\right) \times 1=2^{\dfrac{n(n-1)}{2}}(n \geq 2)\)，

当\(n=1\)时，\(a_1=1\)也符合上式，

\(∴\)数列\(\left\{a_{n}\right\}\)的通项公式为\(a_{n}=2^{\dfrac{n(n-1)}{2}}\)．

(2)\(\because b_{n}=\log _{2} a_{n}=\dfrac{n(n-1)}{2}=\dfrac{n^{2}-n}{2}=\dfrac{1}{2}\left(n^{2}-n\right)\)，

\(\therefore b_{2}+b_{3}+\cdots+b_{100}=\dfrac{1}{2}\left[\left(2^{2}+3^{2}+\cdots+100^{2}\right)-(2+3+\cdots+100)\right]\)

\({\color{Red}{(数列\{b_n\}分成数列\left\{n^{2}\right\}和\left\{n\right\}，再用公式法求解)}}\)

\(=\dfrac{1}{2}\left[\left(1^{2}+2^{2}+3^{2}+\cdots+100^{2}\right)-(1+2+3+\cdots+100)\right]\)

\(=\dfrac{1}{2}\left[\dfrac{100 \times(100+1) \times(2 \times 100+1)}{6}-\dfrac{100 \times(100+1)}{2}\right]\)\(=166650\)

巩固练习

1(★★)已知数列\(\left\{a_{n}\right\}\)的通项\(a_n=2^n+n\)，若数列\(\left\{a_{n}\right\}\)的前\(n\)项和为\(S_n\)，则\(S_8=\)\(\underline{\quad \quad}\)．

2(★★)数列\(1 \dfrac{1}{2}\)，\(2 \dfrac{1}{4}\)，\(3 \dfrac{1}{8}\)，…，\(n+\dfrac{1}{2^{n}}\)的前\(n\)项和为\(S_n=\)\(\underline{\quad \quad}\) 　．

3(★★★)已知数列\(\left\{a_{n}\right\}\)是等比数列，公比为\(q\)，数列\(\{b_n\}\)是等差数列，公差为\(d\)，且满足：\(a_1=b_1=1\)，\(b_2+b_3=4a_2\)，\(a_3-3b_2=-5\)．

(1)求数列\(\left\{a_{n}\right\}\)和\(\{b_n\}\)的通项公式；

(2)设\(c_n=a_n+b_n\)，求数列\(\{c_n\}\)的前\(n\)项和\(S_n\)．

4(★★★)已知公差不为\(0\)的等差数列\(\left\{a_{n}\right\}\)的前9项和\(S_9=45\)，且第\(2\)项、第\(4\)项、第\(8\)项成等比数列．

(1)求数列\(\left\{a_{n}\right\}\)的通项公式；

(2)若数列\(\{b_n\}\)满足\(b_{n}=a_{n}+\left(\dfrac{1}{2}\right)^{n-1}\)，求数列\(\{b_n\}\)的前\(n\)项和\(T_n\)．

参考答案

1.\(546\)

2.\(S_{n}=\dfrac{n(n+1)}{2}-\dfrac{1}{2^{n}}+1\)

3.\((1) a_n=2^{n-1}, b_n=2n-1\)\((2) 2^n+n^2-1\)

4.\((1) a_n=n\)\((2) T_{n}=\dfrac{n^{2}+n+4}{2}-\dfrac{1}{2^{n-1}}\)

【方法四】 错位相减法

当数列\(\left\{a_{n}\right\}\)的通项公式\(a_n=b_n⋅ c_n\),其中\(\{b_n\}\)为等差数列,\(\{c_n\}\)为等比数列.

其解题套路如下

【典题1】已知递增的等比数列\(\left\{a_{n}\right\}\)满足\(a_2+a_3+a_4=28\)，且\(a_3+2\)是\(a_2\),\(a_4\)的等差中项．

(1)求数列\(\left\{a_{n}\right\}\)的通项公式\(a_n\)；

(2)令\(b_{n}=a_{n} \cdot \log _{\frac{1}{2}} a_{n}\)，\(S_{n}=b_{1}+b_{2}+\cdots+b_{n}\)，求\(S_n\)．

【解析】(1)设数列\(\left\{a_{n}\right\}\)的公比为\(q\)，

由题意可知\(\left\{\begin{array}{l}
a_{2}+a_{3}+a_{4}=28 \\
2\left(a_{3}+2\right)=a_{2}+a_{4}
\end{array}\right.\)，

即\(\left\{\begin{array} { l }
{ a _ { 3 } = 8 } \\
{ a _ { 2 } + a _ { 4 } = 2 0 }
\end{array} \Rightarrow \left\{\begin{array}{l}
a_{1} q^{2}=8 \\
a_{1} q+a_{1} q^{3}=20
\end{array}\right.\right.\)，

解得\(\left\{\begin{array}{l}
a_{1}=2 \\
q=2
\end{array}\right.\)或\(\left\{\begin{array}{l}
a_{1}=32 \\
q=\dfrac{1}{2}
\end{array}\right.\)(舍)

\(\therefore a_{n}=2 \cdot 2^{n-1}=2^{n}\).

(2)\(b_{n}=a_{n} \cdot \log _{\frac{1}{2}} a_{n}=2^{n} \cdot \log _{\frac{1}{2}} 2^{n}=-n \cdot 2^{n}\)，

\({\color{Red}{(其中\{n\}是等差数列， \{2^n\}是等比数列，可用错位相减法)}}\)

\(\therefore S_{n}=-1 \times 2-2 \times 2^{2}-3 \times 2^{3}-\cdots-(n-1) \times 2^{n-1}-n \times 2^{n}\)...... (1)

\(2 S_{n}=\quad \quad-1 \times 2^{2}-2 \times 2^{3}-3 \times 2^{4}-\cdots-(n-1) \times 2^{n}-n \times 2^{n+1}\)...... (2)

\(∴(1)-(2)\)得

\(-S_{n}=-\left(2+2^{2}+2^{3}+\cdots 2^{n}\right)+n \times 2^{n+1}\)\(=-\dfrac{2-2^{n+1}}{1-2}+n \times 2^{n+1}\)\(=(n-1) \times 2^{n+1}+2\)

\(\therefore S_{n}=(1-n) \times 2^{n+1}-2\)．

\({\color{Red}{ (最后可用S_2检验运算结果是否正确)}}\)

【典题2】已知正项数列\(\left\{a_{n}\right\}\)的前\(n\)项和为\(S_n\)，满足\(a_n^2+a_n-2S_n=0\)\((n∈N^*)\)．

(1)求数列\(\left\{a_{n}\right\}\)通项公式；

(2)记数列\(\{b_n\}\)的前\(n\)项和为\(S_n\)，若\(b_n=(2a_n-7) 2^n\)，求\(T_n\)；

(3)求数列\(\{T_n\}\)的最小项．

【解析】(1)由\(a_n^2+a_n－2S_n=0\)，

得到\(a_{n+1}^{2}+a_{n+1}-2 S_{n+1}=0\)，

两式相减得\(\left(a_{n+1}^{2}-a_{n}^{2}\right)+\left(a_{n+1}-a_{n}\right)-2\left(S_{n+1}-S_{n}\right)=0\)，

整理得\(\left(a_{n+1}+a_{n}\right)\left(a_{n+1}-a_{n}-1\right)=0\)，

由于数列\(\left\{a_{n}\right\}\)是正项数列，

所以\(a_{n+1}-a_{n}=1\)，

当\(n=1\)时，解得\(a_1=1\)．

故\(a_n=1+n-1=n\)．

(2)由(1)得：\(b_n=(2n-7)⋅2^n\)，

\({\color{Red}{(其中\{2n-7\}是等差数列， \{2^n\}是等比数列，可用错位相减法)}}\)

\(\therefore T_{n}=(-5) \cdot 2^{1}+(-3) \cdot 2^{2}+(-1) \cdot 2^{3}+\cdots+(2 n-9) \cdot 2^{n-1}+(2 n-7) \cdot 2^{n}\)

...... (1) ，

\(2 T_{n}=\quad(-5) \cdot 2^{2}+(-3) \cdot 2^{3}+(-1) \cdot 2^{3}+\cdots+(2 n-9) \cdot 2^{n}+(2 n-7) \cdot 2^{n+1}\)

...... (2)

（1）-（2） 得\(-T_{n}=(-5) \times 2+2^{3}+2^{4}+\cdots+2^{n+1}-(2 n-7) \cdot 2^{n+1}\)，

化简得\(T_{n}=(2 n-9) \cdot 2^{n+1}+18\)．

(3)\(T_{n+1}-T_{n}\)\(=(2 n-7) \cdot 2^{n+2}+18-(2 n-9) \cdot 2^{n+1}-18\)\(=(2 n-5) \cdot 2^{n+1}\)

\({\color{Red}{(做差法判断数列\{T_n\}的单调性，从而求出最小项)}}\)

当\(n≤2\)时，\(T_{n+1}<T_{n}\)，当\(n≥3\)时，\(T_{n+1}>T_{n}\)，

故\(T_1>T_2>T_3<T_4<T_5<⋯\)，

故数列\(\{T_n\}\)的最小值为\(T_3=-30\)．

巩固练习

1(★★★)设等差数列\(\left\{a_{n}\right\}\)的前\(n\)项和为\(S_n\)，且\(S_4=4S_2\)，\(a_{2n}=2a_n+1\)．

(1)求数列\(\left\{a_{n}\right\}\)的通项公式；

(2)设数列\(\{b_n\}\)满足\(b_{n}=\dfrac{2\left(a_{n}-1\right)}{4^{n}}\)，求数列\(\{b_n\}\)的前\(n\)项和\(R_n\)．

2(★★★)正项数列\(\left\{a_{n}\right\}\)的前\(n\)项和为\(S_n\)，且\(8S_n=(a_n+2)^2\)\((n∈N^*)\)．

(1)求\(a_1\)，\(a_2\)的值及数列\(\left\{a_{n}\right\}\)的通项公式；

(2)记\(c_{n}=\dfrac{a_{n}}{3^{n}}\)，数列\(\{c_n\}\)前\(n\)的和为\(T_n\)，求证：\(T_n<2\)．

3(★★★)已知等比数列\(\left\{a_{n}\right\}\)满足\(a_1=2\)，\(a_2=4(a_3－a_4)\)，正项数列\(\{b_n\}\)前\(n\)项和为\(S_n\)，且\(2 \sqrt{S_{n}}=b_{n}+1\)．

(1)求数列\(\left\{a_{n}\right\}\)和\(\{b_n\}\)的通项公式；

(2)令\(c_{n}=\dfrac{b_{n}}{a_{n}}\)，求数列\(\{c_n\}\)的前\(n\)项和\(T_n\)；

(3)若\(λ>0\)，求对所有的正整数\(n\)都有\(2λ^2-kλ+2>a_{2n}b_n\)成立的\(k\)的取值范围．

4(★★★)已知数列\(\left\{a_{n}\right\}\)满足：\((n+1) a_{n+1}-(n+2) a_{n}=(n+1)(n+2)\)\((n∈N^*)\)且\(a_1=4\)，数列\(\{b_n\}\)的前\(n\)项和为\(S_n\)满足：\(S_n=2b_n－1\)\((n∈N^*)\)．

(1)证明数列\(\left\{\dfrac{a_{n}}{n+1}\right\}\)为等差数列，并求数列\(\left\{a_{n}\right\}\)和\(\{b_n\}\)的通项公式；

(2)若\(c_{n}=\left(\sqrt{a_{n}}-1\right) b_{n+1}\)，数列\(\{c_n\}\)的前\(n\)项和为\(T_n\)，对任意的\(n∈N^*\)，\(T_{n} \leq n S_{n+1}-m-2\)恒成立，求实数\(m\)的取值范围．

参考答案

1. \((1) a_n=2n-1\)\((2) R_{n}=\dfrac{1}{9}\left(4-\dfrac{3 n+1}{4^{n-1}}\right)\)

2. \((1) a_n=4n－2\)\((2) T_{n}=2-\dfrac{2 n+2}{3^{n}}<2\)

3. \((1)a_{n}=\dfrac{1}{2^{n-2}},b_{n}=2 n-1\)
   
   \((2)T_{n}=\dfrac{3}{2}+(2 n-3) \cdot 2^{n-1}\)
   
   \((3)(-\infty, 2 \sqrt{2})\)

4. \((1) a_{n}=(n+1)^{2}, b_{n}=2^{n-1}\)
   
   \((2) m≤－1\)

【方法五】 裂项相消法

常见裂项公式

\((1)\dfrac{1}{n(n+1)}=\dfrac{1}{n}-\dfrac{1}{n+1}\)，\(\dfrac{1}{n(n+k)}=\dfrac{1}{k}\left(\dfrac{1}{n}-\dfrac{1}{n+k}\right)\)

\((2)\dfrac{1}{\sqrt{n+1}+\sqrt{n}}=\sqrt{n+1}-\sqrt{n}\)，\(\dfrac{1}{\sqrt{n+k}+\sqrt{n}}=\dfrac{1}{k}(\sqrt{n+k}-\sqrt{n})\)

【典题1】设等差数列\(\left\{a_{n}\right\}\)满足\(a_2=5\)，\(a_6+a_8=30\)，则数列\(\left\{\dfrac{1}{a_{n}^{2}-1}\right\}\)的前\(n\)项的和等于\(\underline{\quad \quad}\)．

【解析】\(∵a_6+a_8=30\)，\(∴a_7=15\)，

又\(∵a_2=5\)，\(\therefore d=\dfrac{15-5}{7-2}=2\)，

\(∴a_n=a_2+(n-2)d=2n+1\)，

\(∴a_n^2=(2n+1)^2=4n^2+4n+1\)，

\(\therefore \dfrac{1}{a_{n}^{2}-1}=\dfrac{1}{4 n^{2}+4 n}=\dfrac{1}{4}\left(\dfrac{1}{n}-\dfrac{1}{n+1}\right)\)

\({\color{Red}{ (因式分解裂项是关键) }}\)

\(∴\)数列\(\left\{\dfrac{1}{a_{n}^{2}-1}\right\}\)的前\(n\)项的和为：

\(\dfrac{1}{4}\left[\left(1-\dfrac{1}{2}\right)+\left(\dfrac{1}{2}-\dfrac{1}{3}\right)+\cdots+\left(\dfrac{1}{n}-\dfrac{1}{n+1}\right)\right]\)\(=\dfrac{1}{4}\left[1-\dfrac{1}{n+1}\right]=\dfrac{n}{4(n+1)}\).

【点拨】

本题是用了常见的裂项公式\(\dfrac{1}{n(n+1)}=\dfrac{1}{n}-\dfrac{1}{n+1}\)，\(\dfrac{1}{n(n+k)}=\dfrac{1}{k}\left(\dfrac{1}{n}-\dfrac{1}{n+k}\right)\)，

思考下以下各项怎么裂项：\(a_{n}=\dfrac{1}{n^{2}-n}(n \geq 2)\)，\(a_{n}=\dfrac{1}{2 n^{2}+4 n}\)，\(a_{n}=\dfrac{1}{n^{2}+3 n}\).

【典题2】数列\(\left\{a_{n}\right\}\)的通项公式\(a_{n}=\dfrac{1}{\sqrt{n+2}+\sqrt{n+3}}\)，则该数列的前\(n\)项和为\(S_n\)等于\(\underline{\quad \quad}\).

【解析】\(a_{n}=\dfrac{1}{\sqrt{n+2}+\sqrt{n+3}}\)\(=\dfrac{\sqrt{n+3}-\sqrt{n+2}}{(\sqrt{n+2}+\sqrt{n+3})(\sqrt{n+3}-\sqrt{n+2})}\)\(=\dfrac{\sqrt{n+3}-\sqrt{n+2}}{(n+3)-(n+2)}=\sqrt{n+3}-\sqrt{n+2}\)

\(\begin{aligned}
\therefore S_{n}=& a_{1}+a_{2}+a_{3}+\cdots+a_{n-1}+a_{n} \\
=&(\sqrt{4}-\sqrt{3})+(\sqrt{5}-\sqrt{4})+(\sqrt{6}-\sqrt{5})+\cdots \\
&+(\sqrt{n+2}-\sqrt{n+1})+(\sqrt{n+3}-\sqrt{n+2}) \\
=& \sqrt{n+3}-\sqrt{3} .
\end{aligned}\)

【点拨】

① 本题是用了常见的裂项公式\(\dfrac{1}{\sqrt{n+k}+\sqrt{n}}=\dfrac{1}{k}(\sqrt{n+k}-\sqrt{n})\)，有些类似分母有理化，\(\sqrt{n+k}+\sqrt{n+b}\)与\(\sqrt{n+k}-\sqrt{n+b}\)互为“共轭根式”.

② 思考下以下各项怎么裂项：\(a_{n}=\dfrac{1}{\sqrt{n+1}-\sqrt{n}}\)，\(a_{n}=\dfrac{1}{\sqrt{n+3}-\sqrt{n}}\)，\(a_{n}=\dfrac{1}{\sqrt{n+1}-\sqrt{n-1}}\).

【典题3】等比数列\(\left\{a_{n}\right\}\)中，\(a_1=2\)，\(q=2\)，数列\(b_{n}=\dfrac{a_{n+1}}{\left(a_{n+1}-1\right)\left(a_{n}-1\right)}\)，\(\{b_n\}\)的前\(n\)项和为\(T_n\)，则\(T_{10}\)的值为\(\underline{\quad \quad}\)．

【解析】由题意，可知\(a_{n}=2 \times 2^{n-1}=2^{n}\),\(n∈N^*\)

则\(b_{n}=\dfrac{a_{n+1}}{\left(a_{n+1}-1\right)\left(a_{n}-1\right)}=\dfrac{2^{n+1}}{\left(2^{n+1}-1\right)\left(2^{n}-1\right)}\)\(=2\left(\dfrac{1}{2^{n}-1}-\dfrac{1}{2^{n+1}-1}\right)\)

\(\begin{aligned}
\therefore T_{10}=& b_{1}+b_{2}+\cdots+b_{10} \\
=& 2\left(\dfrac{1}{2^{1}-1}-\dfrac{1}{2^{2}-1}+\dfrac{1}{2^{2}-1}-\dfrac{1}{2^{3}-1}+\cdots+\dfrac{1}{2^{10}-1}-\dfrac{1}{2^{11}-1}\right) \\
=& 2\left(\dfrac{1}{2^{1}-1}-\dfrac{1}{2^{11}-1}\right) \\
=& \dfrac{4092}{2047} .
\end{aligned}\)．

【点拨】

① 本题的裂项\(\dfrac{2^{n+1}}{\left(2^{n+1}-1\right)\left(2^{n}-1\right)}=2\left(\dfrac{1}{2^{n}-1}-\dfrac{1}{2^{n+1}-1}\right)\)需要一些技巧，可这么猜想\(\dfrac{2^{n+1}}{\left(2^{n+1}-1\right)\left(2^{n}-1\right)}\)中分母有\(2^{n+1}-1\)与\(2^n-1\)，往裂项的角度思考，那它是否等于\(\dfrac{1}{2^{n}-1}-\dfrac{1}{2^{n+1}-1}\)(当然是分母小的减去分母大)呢？我们就看下\(\dfrac{1}{2^{n}-1}-\dfrac{1}{2^{n+1}-1}\)通分后的结果\(\dfrac{2^{n}}{\left(2^{n+1}-1\right)\left(2^{n}-1\right)}\)与“目标\(\dfrac{2^{n+1}}{\left(2^{n+1}-1\right)\left(2^{n}-1\right)}\)”不相等，但是\(2\)倍的关系，故可得\(\dfrac{2^{n+1}}{\left(2^{n+1}-1\right)\left(2^{n}-1\right)}=2\left(\dfrac{1}{2^{n}-1}-\dfrac{1}{2^{n+1}-1}\right)\)

② 在裂项的技巧中，大胆猜想再小心验证便可.

思考下以下各项怎么裂项：\(a_{n}=\dfrac{2^{n-1}}{\left(2^{n-1}+1\right)\left(2^{n}+1\right)}\)，\(a_{n}=\dfrac{2 \cdot 3^{n-1}}{\left(2 \cdot 3^{n-1}+2\right)\left(2 \cdot 3^{n}+2\right)}\).

【典题4】已知数列\(\left\{a_{n}\right\}\)满足\(a_n≠0\)，\(a_{1}=\dfrac{1}{3}\)，\(a_{n-1}-a_{n}=2 a_{n} a_{n-1}\)\((n≥2 ,n∈N^*)\)．

(1)求证：\(\left\{\dfrac{1}{a_{n}}\right\}\)是等差数列；

(2)证明：\(a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}<\dfrac{1}{4}\)．

【解析】证明：(1)\(\because a_{n-1}-a_{n}=2 a_{n} a_{n-1}\)\((n≥2 ,n∈N^*)\)

\(\therefore \dfrac{1}{a_{n}}-\dfrac{1}{a_{n-1}}=2(n \geq 2)\)

\(\therefore\left\{\dfrac{1}{a_{n}}\right\}\)是以3为首项，\(2\)为公差的等差数列．

(2)由(1)知：\(\dfrac{1}{a_{n}}=3+(n-1) \cdot 2=2 n+1\)

\(\therefore a_{n}=\dfrac{1}{2 n+1}\)

\(\therefore a_{n}^{2}=\dfrac{1}{(2 n+1)^{2}}\)\(<\dfrac{1}{4 n^{2}+4 n}=\dfrac{1}{4 n(n+1)}\)\(=\dfrac{1}{4}\left(\dfrac{1}{n}-\dfrac{1}{n+1}\right)\) \({\color{Red}{(放缩法) }}\)

\(\begin{aligned}
&\therefore a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2} \\
&<\dfrac{1}{4}\left(\dfrac{1}{1}-\dfrac{1}{2}\right)+\dfrac{1}{4}\left(\dfrac{1}{2}-\dfrac{1}{3}\right)+\cdots+\dfrac{1}{4}\left(\dfrac{1}{n}-\dfrac{1}{n+1}\right) \\
&<\dfrac{1}{4}\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\cdots+\dfrac{1}{n}-\dfrac{1}{n+1}\right) \\
&=\dfrac{1}{4}\left(1-\dfrac{1}{n+1}\right)<\dfrac{1}{4}
\end{aligned}\)

【点拨】

① 在数列中求证不等式，利用放缩法是常用的方法，但技巧性较高.

② 要证明\(a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}<\dfrac{1}{4}\)，用到放缩法的话，可考虑把\(a_{n}^{2}=\dfrac{1}{(2 n+1)^{2}}\)“放大些”，则要把分母\((2 n+1)^{2}=4 n^{2}+4 n+1\)“缩小些”，缩小多少呢？那“消掉”常数项1，对\(4n^2+4n+1\)来说“影响较小”，并且\(a_{n}^{2}=\dfrac{1}{(2 n+1)^{2}}<\dfrac{1}{4 n^{2}+4 n}=\dfrac{1}{4 n(n+1)}\)\(=\dfrac{1}{4}\left(\dfrac{1}{n}-\dfrac{1}{n+1}\right)\)放缩后还能裂项求和.

巩固练习

1(★★)数列\(\left\{a_{n}\right\}\)满足\(a_{n}=\dfrac{1}{(2 n+1)(2 n+3)}\),\(n∈N^*\)，其前\(n\)项和为\(S_n\)．若\(S_n<M\)恒成立，则\(M\)的最小值为\(\underline{\quad \quad}\)．

2(★★★)已知正项数列\(\left\{a_{n}\right\}\)的前\(n\)项和为\(S_n\)，对\(∀n∈N^*\)有\(2S_n=a_n^2+a_n\)．令\(b_{n}=\dfrac{1}{a_{n} \sqrt{a_{n+1}}+a_{n+1} \sqrt{a_{n}}}\)，设\(\left\{b_{n}\right\}\)的前\(n\)项和为\(T_n\)，则在\(T_1\),\(T_2\),\(T_3\),… ,\(T_{100}\)中有理数的个数为\(\underline{\quad \quad }\)．

3(★★★)已知数列\(\left\{a_{n}\right\}\)的前\(n\)项和为\(S_n\)，且满足\(a_1=2\)，\(S_{n}=a_{n+1}-2^{n+2}+2\),\(n∈N^*\)．

(1)求数列\(\left\{a_{n}\right\}\)的通项公式；

(2)设\(b_{n}=\dfrac{2^{n}}{a_{n}}\)，记数列\(\left\{b_{n} b_{n+1}\right\}\)的前\(n\)项和为\(T_n\)，证明：\(\dfrac{1}{2} \leq T_{n}<1\)．

4(★★★)已知数列\(\left\{a_{n}\right\}\)满足\(a_1=1\)，\(a_{n+1}=\dfrac{a_{n}}{a_{n}+1}\)．

(1)证明：数列\(\left\{\dfrac{1}{a_{n}}\right\}\)是等差数列，并求数列\(\{a_n\}\)的通项公式；

(2)设\(b_{n}=\dfrac{a_{n}}{n+2}\)，求数列\(\{b_n\}\)前\(n\)项和\(S_n\)．

5(★★★)设数列\(\left\{a_{n}\right\}\)的前\(n\)项和为\(S_n\)，已知\(a_n>0\)，\(a_n^2+2a_n=4S_n+3\)．

(1)求\(\{a_n\}\)的通项公式；

(2)若数列\(\{b_n\}\)满足\(b_{n}=\dfrac{2 n+1}{n^{2}\left(a_{n+1}-1\right)^{2}}\)，求\(\{b_n\}\)的前\(n\)项和\(T_n\)．

6(★★★★)设\(S_n\)为数列\(\left\{a_{n}\right\}\)的前\(n\)项和，且\(S_{n+1}=3 S_{n}+4 n\)\((n∈N^*)\)，\(a_1=0\)．

(1)求证：数列\(\left\{a_{n}+2\right\}\)是等比数列；

(2)若对任意\(T_n\)为数列\(\left\{\dfrac{a_{n}+2}{\left(a_{n}+4\right)\left(a_{n+1}+4\right)}\right\}\)的前\(n\)项和，求证：\(T_{n}<\dfrac{1}{2}\)．

7(★★★★)已知数列\(\left\{a_{n}\right\}\)的前\(n\)项和为\(S_n\)，已知\(a_1=2\)，\(6 S_{n}=3 n a_{n+1}-2n(n+1)(n+2)\)，\(n∈N^*\)．

(1)求数列\(\left\{a_{n}\right\}\)的通项公式；

(2)证明：\(\dfrac{1}{a_{1}}+\dfrac{1}{a_{2}}+\cdots+\dfrac{1}{a_{n}}<\dfrac{5}{6}\)．

参考答案

1. \(\dfrac{1}{6}\)

2. \(9\)

3. \((1)a_n=n\cdot 2^n\)
   
   \((2)\)提示\(T_{n}=1-\dfrac{1}{n+1}\)

4. \((1)a_{n}=\dfrac{1}{n}\)
   
   \((2)S_{n}=\dfrac{3}{4}-\dfrac{2 n+3}{2(n+1)(n+2)}\)

5. \((1)a_n=2n+1\)
   
   \((2)T_{n}=\dfrac{n^{2}+2 n}{4(n+1)^{2}}\)

6. \((1)\)提示：定义法证明
   
   \((2)\)提示：裂项相消法求\(T_{n}=\dfrac{1}{4}\left(\dfrac{1}{2}-\dfrac{1}{3^{n}+1}\right)\)

7. \((1)a_n=2n^2\)
   
   \((2)\)提示：放缩法、裂项相消法