数列专题2 求数列的前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
求数列的前\(n\)项和\(S_n\)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.
基本方法
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\(3+6+12+⋯+3\cdot 2^{n-2}\).
解析 和式\(3+6+12+⋯+3\cdot 2^{n-2}\)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\(3+6+12+⋯+3\cdot 2^{n-2}\)只有\(n-1\)项而已,
则\(3+6+12+⋯+3\cdot 2^{n-2}\) (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.
巩固练习
1.求和式\(1+4+7+⋯+(3n+1)\).
2.已知\(\{a_n\}\)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\(S_n\).
参考答案
-
答案 \(\dfrac{3 n^2+5 n+2}{2}\)
解析 \(1+4+7+⋯+(3n+1)=\dfrac{3 n^2+5 n+2}{2}\). -
答案 \(S_n=2^{n+1}-2\)
解析 \(\because\)数列\(\{a_n\}\)是等差数列,公差\(d≠0\),\(a_1=1\),且\(a_1\) ,\(a_3\) ,\(a_9\)成等比数列,
\(\therefore (1+2d)^2=1×(1+8d)\),解得\(d=1\)或\(d=0\)(舍),
\(\therefore a_n=a_1+(n-1)d=n\),
\(\therefore 2^{a_n}=2^n\),
\(\therefore\)数列\(\left\{2^{a_n}\right\}\)是首项为\(2\),公比为\(2\)的等比数列,
\(\therefore S_n=\dfrac{2\left(1-2^n\right)}{1-2}=2^{n+1}-2\).
方法二 倒序相加(乘)法
1 对于某个数列\(\{a_n\}\),若满足\(a_1+a_n=a_2+a_{n-1}=⋯=a_k+a_{n-k+1}\),则求前\(n\)项和\(S_n\)可使用倒序相加法.
具体解法 设\(S_n=a_1+a_2+⋯+a_{n-1}+a_n\) ①
把①反序可得\(S_n=a_n +a_{n-1}+⋯+a_2+a_1\) 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\(\{a_n\}\),若满足\(a_1+a_n=a_2+a_{n-1}=⋯=a_k+a_{n-k+1}\),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\(f(-3)+f(4)= \dfrac{1}{2}\),\(f(-2)+f(3)= \dfrac{1}{2}\),\(f(-1)+f(2)= \dfrac{1}{2}\),\(f(0)+f(1)= \dfrac{1}{2}\)
\(f(-3)+f(-2)+⋯+f(0)+⋯+f(3)+f(4)=4× \dfrac{1}{2}=2\).
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.
巩固练习
1.设等差数列\(\{a_n\}\),公差为\(d\),求证:\(\{a_n\}\)的前\(n\)项和 \(S_n=\dfrac{\left(a_1+a_n\right) n}{2}\).
2.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
-
证明 \(S_n=a_1+a_2+a_3+⋯+a_n\) ①
倒序得:\(S_n=a_n+a_{n-1}+a_{n-2}+⋯+a_1\) ②
①+②得:\(2S_n=(a_1+a_n)+(a_2+a_{n-1})+(a_3+a_{n-2})+⋯+(a_n+a_1)\)
又\(\because a_1+a_n=a_2+a_{n-1}=a_3+a_{n-2}=⋯=a_n+a_1\)
\(\therefore 2 S_n=n\left(a_1+a_n\right) \Rightarrow S_n=\dfrac{n\left(a_1+a_n\right)}{2}\). -
答案 \(\dfrac{7}{2}\)
解析 \(\because\)函数 \(f(x)=\dfrac{x^2}{1+x^2}\),
\(\therefore f(x)+f\left(\dfrac{1}{x}\right)=\dfrac{x^2}{1+x^2}+\dfrac{\dfrac{1}{x^2}}{1+\dfrac{1}{x^2}}=\dfrac{x^2}{1+x^2}+\dfrac{1}{x^2+1}=1\),
\(\therefore f(1)+f(2)+f\left(\dfrac{1}{2}\right)+f(3)+f\left(\dfrac{1}{3}\right)+f(4)+f\left(\dfrac{1}{4}\right)\)\(=f(1)+1+1+1=\dfrac{1}{1+1}+3=\dfrac{7}{2}\).
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\(\{a_n\}\),\(\{b_n\}\)之和,则数列\(\{c_n\}\)的前\(n\)项和等于两个数列\(\{a_n\}\),\(\{b_n\}\)的前\(n\)项和的和.
2 常见的是\(c_n=\)等差+等比的形式,
![](https://cdn.nlark.com/yuque/0/2022/png/21447551/1669463412147-f555001f-99d5-431a-9944-7f2efa8a6008.png)
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\(\{3^n+2n-1\}\)的前\(n\)项和\(S_n\).
解析 设\(a_n=3^n+2n-1\), (其中可知数列\(\{3^n\}\)是等比数列,数列\(\{2n-1\}\)是等差数列)
则\(S_n=a_1+a_2+a_3+⋯+a_n\)\(=(3^1+1)+(3^2+3)+(3^3+5)+⋯(3^n+2n-1)\)
(把等比项和等差项分别放在一组)
\(=(3^1+3^2+3^3+⋯+3^n )+(1+3+5+⋯+2n-1)\) (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】 已知在等差数列\(\{a_n\}\)中,\(a_1=2\),\(a_3+a_5=10\).
(1)设 \(b_n=2^{a_n}\),求证:数列\(\{b_n\}\)是等比数列;(2)求数列\(\{a_n+b_n\}\)的前\(n\)项和.
解析 (1)设公差为\(d\)的等差数列\(\{a_n\}\)中,\(a_1=2\),\(a_3+a_5=10\).
整理得 \(\left\{\begin{array}{c}
a_1=2 \\
2 a_1+6 d=10
\end{array}\right.\),解得 \(\left\{\begin{array}{l}
a_1=2 \\
d=1
\end{array}\right.\),
所以\(a_n=a_1+(n-1)=n+1\).
由于 \(b_n=2^{a_n}\),所以\(b_n=2^{n+1}\),\(b_{n-1}=2^n\),
整理得 \(\dfrac{b_n}{b_{n-1}}=2\)(常数),
所以数列\(\{b_n\}\)是以\(b_1=2^2=4\)为首项,\(2\)为公比的等比数列.
(2)由于数列\(\{b_n\}\)是以\(b_1=2^2=4\)为首项,\(2\)为公比的等比数列,
所以\(b_n=4×2^{n-1}=2^{n+1}\).
所以\(a_n+b_n=2^{n+1}+n+1\),
故\(S_n=\dfrac{4\left(2^n-1\right)}{2-1}+\dfrac{n(2+n+1)}{2}=2^{n+2}+\dfrac{n(n+3)}{2}-4\).
巩固练习
1.已知数列\(\{a_n\}\)的通项\(a_n=2^n+n\),若数列\(\{a_n\}\)的前\(n\)项和为\(S_n\),则\(S_8=\) \(\underline{\quad \quad}\).
2.已知数列\(\{a_n\}\)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\(\{a_n\}\)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\(S_n\).
参考答案
-
答案 \(546\)
解析 数列\(\{a_n\}\)的通项\(a_n=2^n+n\),
若数列\(\{a_n\}\)的前\(n\)项和为\(S_n\),
则 \(S_n=\left(2^1+2^2+\cdots+2^n\right)+(1+2+\cdots+n)=\dfrac{2\left(2^n-1\right)}{2-1}+\dfrac{n(n+1)}{2}\)\(=2\left(2^n-1\right)+\dfrac{n(n+1)}{2}\).
则\(S_8=2\left(2^8-1\right)+\dfrac{8 \times 9}{2}=546\) . -
答案 (1) \(a_n=2^{n-1}\),\(b_n=2n-1\) ;(2) \(2^n+n^2-1\)
解析 (1)设等比数列\(\{a_n\}\)的公比为\(q\),等差数列\(\{b_n\}\)的公差为\(d\),
由题意知\(q>0\),
由已知,有 \(\left\{\begin{array}{l} (1+d)+(1+2 d)=4 q \\ q^2-3(1+d)=-5 \end{array}\right.\),
即 \(\left\{\begin{array}{l} -4 q+3 d=-2 \\ q^2-3 d=-2 \end{array}\right.\),解得\(q=d=2\).
\(\therefore \{a_n\}\)的通项公式为\(a_n=2^{n-1}\),\(\{b_n\}\)的通项公式为\(b_n=2n-1\);
(2)由(1)知,\(c_n=a_n+b_n=2^{n-1}+2n-1\),
则\(S_n=(1+2^1+2^2+⋯+2^{n-1})+[1+3+5+⋯+(2n-1)]\)
\(=\dfrac{1 \times\left(1-2^n\right)}{1-2}+\dfrac{n(1+2 n-1)}{2}=2^n+n^2-1\).
方法四 错位相减法
当数列\(\{a_n\}\) 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.
举例:
【典题1】 已知递增的等比数列\(\{a_n\}\)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\(\{a_n\}\)的通项公式\(a_n\);
(2)令\(b_n=a_n⋅\log _{\dfrac{1}{2}} a_n\),\(S_n=b_1+b_2+⋯+b_n\),求\(S_n\).
解析 (1)设数列\(\{a_n\}\)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)
\(\therefore a_n=2⋅2^{n-1}=2^n\).
(2)\(b_n=a_n\cdot \log _{\dfrac{1}{2}} a_n=2^n\cdot \log _{\dfrac{1}{2}} 2^n=-n⋅2^n\),
(其中\(\{n\}\)是等差数列,\(\{2^n\}\)是等比数列,可用错位相减法)
\(\therefore S_n=-1×2-2×2^2-3×2^3-⋯-{n-1}×2^{n-1}-n×2^n\) (1)
\(2S_n=\) \(-1×2^2-2×2^3-3×2^4-⋯-{n-1}×2^n-n×2^{n+1}\) (2)
\(\therefore (1)-(2)\)得
\(-S_n=-\left(2+2^2+2^3+\cdots 2^n\right)+n \times 2^{n+1}\)\(=-\dfrac{2-2^{n+1}}{1-2}+n \times 2^{n+1}=(n-1) \times 2^{n+1}+2\)
\(\therefore S_n=(1-n)×2^{n+1}-2\). (最后可用\(S_2\)检验运算结果是否正确)
巩固练习
1.设等差数列\(\{a_n\}\)的前\(n\)项和为\(S_n\),且\(S_4=4S_2\),\(a_{2n}=2a_n+1\).
(1)求数列\(\{a_n\}\)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.已知正项数列\(\{a_n\}\)的前\(n\)项和为\(S_n\),满足\(a_n^2+a_n-2S_n=0(n∈N^*)\).
(1)求数列\(\{a_n\}\)的通项公式;
(2)记数列\(\{b_n\}\)的前\(n\)项和为\(T_n\),若\(b_n=(2a_n-7) 2^n\),求\(T_n\);
(3)求数列\(\{T_n\}\)的最小项.
参考答案
-
答案 (1)\(a_n=2n-1\); (2) \(R_n=\dfrac{1}{9}\left(4-\dfrac{3 n+1}{4^{n-1}}\right)\).
解析 (1)设等差数列\(\{a_n\}\)的首项为\(a_1\),公差为\(d\),
由\(S_4=4S_2\),\(a_{2n}=2a_n+1\)得\(\left\{\begin{array}{l} 4 a_1+6 d=8 a_1+4 d \\ a_1+(2 n-1) d=2 a_1+2(n-1) d+1 \end{array}\right.\),
解得\(a_1=1\),\(d=2\).
因此\(a_n=2n-1(n∈N^*)\).
(2)由题意知: \(b_n=\dfrac{2 a_n-2}{4^n}=(n-1)\left(\dfrac{1}{4}\right)^{n-1}\),
所以 \(R_n=0 \times\left(\dfrac{1}{4}\right)^0+1 \times\left(\dfrac{1}{4}\right)^1+2 \times\left(\dfrac{1}{4}\right)^2+3 \times\left(\dfrac{1}{4}\right)^3+\cdots+(n-1) \times\left(\dfrac{1}{4}\right)^{n-1}\),
\(\dfrac{1}{4} R_n=0 \times\left(\dfrac{1}{4}\right)^1+1 \times\left(\dfrac{1}{4}\right)^2+2 \times\left(\dfrac{1}{4}\right)^3+\cdots+(n-2) \times\left(\dfrac{1}{4}\right)^{n-1}+(n-1) \times\left(\dfrac{1}{4}\right)^n\)
两式相减得 \(\dfrac{3}{4} R_n=\dfrac{1}{4}+\left(\dfrac{1}{4}\right)^2+\cdots+\left(\dfrac{1}{4}\right)^{n-1}-(n-1)\left(\dfrac{1}{4}\right)^n=\dfrac{\dfrac{1}{4}-\left(\dfrac{1}{4}\right)^n}{1-\dfrac{1}{4}}-(n-1)\left(\dfrac{1}{4}\right)^n\)
整理得\(R_n=\dfrac{1}{9}\left(4-\dfrac{3 n+1}{4^{n-1}}\right)\),
所以数列\(\{c_n\}\)的前\(n\)项和\(R_n=\dfrac{1}{9}\left(4-\dfrac{3 n+1}{4^{n-1}}\right)\). -
答案 (1)\(a_n=n\);(2)\(T_n=(2n-9)⋅2^{n+1}+18\);(3) \(-30\).
解析 (1)由\(a_n^2+a_n-2S_n=0\),得到:\(a_{n+1}^2+a_{n+1}-2S_{n+1}=0\),
两式相减得:\((a_{n+1}^2-a_n^2)+(a_{n+1}-a_n)-2(S_{n+1}-S_n)=0\),
整理得:\((a_{n+1}+a_n)(a_{n+1}-a_n-1)=0\),
由于数列\(\{a_n\}\)是正项数列,
所以\(a_{n+1}-a_n=1\)(常数),
当\(n=1\)时,解得\(a_1=1\).
故\(a_n=1+n-1=n\).
(2)由(1)得:\(b_n=(2n-7)⋅2^n\),
所以:\(T_n=(-5)⋅2^1+(-3)⋅2^2+(-1)⋅2^3+⋯+(2n-7)⋅2^n\)①,
\(2T_n=(-5)⋅2^2+(-3)⋅2^3+(-1)⋅2^3+⋯+(2n-7)⋅2^{n+1}\)②,
①-②得:\(-T_n=(-5)×2+2^3+2^4+⋯+2^{n+1}-(2n-7)⋅2^{n+1}\),
解得:\(T_n=(2n-9)⋅2^{n+1}+18\).
(3)\(T_{n+1}-T_n=(2n-7)⋅2^{n+2}+18-(2n-9)⋅2^{n+1}-18=(2n-5)⋅2^{n+1}\),
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\(T_3=-30\).
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\(\{a_n\}\)满足\(a_2=5\),\(a_6+a_8=30\),则数列\(\left\{\dfrac{4}{a_n^2-1}\right\}\)的前\(n\)项的和等于\(\underline{\quad \quad}\).
解析 \(\because a_6+a_8=30\), \(\therefore a_7=15\),
又\(\because a_2=5\), \(\therefore d=\dfrac{15-5}{7-2}=2\),
\(\therefore a_n=a_2+{n-2}d=2n+1\),
\(\therefore a_n^2=(2n+1)^2=4n^2+4n+1\),
\(\therefore \dfrac{4}{a_n^2-1}=\dfrac{4}{4 n^2+4 n}=\dfrac{1}{n^2+n}=\dfrac{1}{n(n+1)}=\dfrac{1}{n}-\dfrac{1}{n+1}\) (因式分解裂项是关键)
\(\therefore\) 数列 \(\left\{\dfrac{4}{a_n^2-1}\right\}\)的前\(n\)项的和为 \(\left(1-\dfrac{1}{2}\right)+\left(\dfrac{1}{2}-\dfrac{1}{3}\right)+\cdots+\left(\dfrac{1}{n}-\dfrac{1}{n+1}\right)=1-\dfrac{1}{n+1}\).
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】 已知数列\(\{a_n\}\)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\(\{a_n\}\)的通项公式;
(2)设 \(b_n=\dfrac{2^n}{a_n}\),记数列\(\{b_n b_{n+1}\}\)的前\(n\)项和为\(T_n\),证明:\(\dfrac{1}{2}≤T_n<1\).
解析 (1)解:由题意,当\(n=1\)时,\(a_1=S_1=a_2-2^3+2\),
即\(a_2-6=2\),解得\(a_2=8\),
当\(n≥2\)时,由\(S_n=a_{n+1}-2^{n+2}+2\),可得:\(S_{n-1}=a_n-2^{n+1}+2\),
两式相减,可得:\(a_n=a_{n+1}-a_n-2^{n+2}-2^{n+1}\),
整理,得\(a_{n+1}-2a_n=2^{n+1}\),
两边同时乘以\(\dfrac{1}{2^{n+1} }\),可得\(\dfrac{a_{n+1}}{2^{n+1}}-\dfrac{a_n}{2^n}=1\),\((n≥2)\)
\(\because \dfrac{a_ 1}{2^1 }=1\), \(\therefore\)数列\(\left\{\dfrac{a_n}{2^n}\right\}\)是以\(1\)为首项,\(1\)为公差的等差数列,
\(\therefore \dfrac{a_n}{2^n}=1+1 \times(n-1)=n\),\(\therefore a_n=n2^n\),\(n∈N^*\).
(2)证明:由(1)知,\(b_n=\dfrac{2^n}{a_n}=\dfrac{2^n}{n \cdot 2^n}=\dfrac{1}{n}\) ,
则\(b_n b_{n+1}=\dfrac{1}{n(n+1)}=\dfrac{1}{n}-\dfrac{1}{n+1}\),
\(\therefore T_n=b_1 b_2+b_2 b_3+\cdots+b_n b_{n+1}=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\cdots+\dfrac{1}{n}-\dfrac{1}{n+1}=1-\dfrac{1}{n+1}\),
\(\because n∈N^*\), \(\therefore 0<\dfrac{1}{n+1} \leq \dfrac{1}{2}\),
\(\therefore \dfrac{1}{2} \leq 1-\dfrac{1}{n+1}<1\),即\(\dfrac{1}{2}≤T_n<1\).
巩固练习
1.已知数列\(\{a_n\}\)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\(S_n\).
2.已知正项数列\(\{a_n\}\)的前\(n\)项和为\(S_n\),\(4S_n=a_n^2+4n-1\),\(a_1=1\).
(1)求数列\(\{a_n\}\)的通项公式;
(2)设\(\{a_n\}\)是递增数列,\(b_n=\dfrac{1}{a_n \cdot a_{n+1}}\),\(T_n\)为数列\(\{b_n\}\)的前\(n\)项和,若\(T_n \leq \dfrac{m}{6}\)恒成立,求实数\(m\)的取值范围.
参考答案
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答案 (1)\(a_n=\dfrac{1}{n}\); (2) \(S_n=\dfrac{3}{4}-\dfrac{2 n+3}{2(n+1)(n+2)}\)
解析 (1)证明:由 \(a_{n+1}=\dfrac{a_n}{a_n+1}\),得 \(\dfrac{1}{a_{n+1}}-\dfrac{1}{a_n}=\dfrac{a_n+1}{a_n}-\dfrac{1}{a_n}=1\),
再由\(a_1=1\),得 \(\dfrac{1}{a_1}=1\),
\(\therefore\)数列\(\left\{\dfrac{1}{a_n}\right\}\)是首项为\(1\),公差为\(1\)的等差数列,
\(\therefore \dfrac{1}{a_n}=1+(n-1) \times 1=n\),则\(a_n=\dfrac{1}{n}\);
(2)解:由 \(b_n=\dfrac{a_n}{n+2}\),得 \(b_n=\dfrac{1}{n(n+2)}=\dfrac{1}{2}\left(\dfrac{1}{n}-\dfrac{1}{n+2}\right)\),
\(\therefore\) 数列\(\{b_n\}\)前\(n\)项和 \(S_n=\dfrac{1}{2}\left[\left(1-\dfrac{1}{3}\right)+\left(\dfrac{1}{2}-\dfrac{1}{4}\right)+\left(\dfrac{1}{3}-\dfrac{1}{5}\right)+\cdots+\left(\dfrac{1}{n-1}-\dfrac{1}{n+1}\right)+\left(\dfrac{1}{n}-\dfrac{1}{n+2}\right)\right]\)
\(=\dfrac{1}{2}\left(1+\dfrac{1}{2}-\dfrac{1}{n+1}-\dfrac{1}{n+2}\right)=\dfrac{3}{4}-\dfrac{2 n+3}{2(n+1)(n+2)}\). -
答案 (1) \(a_n=2n-1\)或\(a_n=1\); (2) \([3,+∞)\).
解析 (1)\(n≥2\)时,\(4a_n=4S_n-4S_{n-1}=a_n^2+4n-1-[a_{n-1}^2+4{n-1}-1]\),
化为:\((a_n-2)^2=a_{n-1}^2\),\(a_n>0\).
\(\therefore a_n-a_{n-1}=2\),或\(a_n+a_{n-1}=2\),
\(a_n-a_{n-1}=2\)时,数列\(\{a_n\}\)是等差数列,\(a_n=1+2(n-1)=2n-1\).
\(a_n+a_{n-1}=2\)时,\(\because a_1=1\),可得\(a_n=1\).
(2) \(\{a_n\}\)是递增数列,\(\therefore a_n=2n-1\).
\(b_n=\dfrac{1}{a_n \cdot a_{n+1}}=\dfrac{1}{(2 n-1)(2 n+1)}=\dfrac{1}{2}\left(\dfrac{1}{2 n-1}-\dfrac{1}{2 n+1}\right)\),
数列\(\{b_n\}\)的前\(n\)项和 \(T_n=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\cdots \cdots+\dfrac{1}{2 n-1}-\dfrac{1}{2 n+1}\right)=\dfrac{1}{2}\left(1-\dfrac{1}{2 n+1}\right)<\dfrac{1}{2}\),
\(\because T_n \leq \dfrac{m}{6}\)恒成立, \(\therefore \dfrac{1}{2} \leq \dfrac{m}{6}\),解得\(m≥3\).
\(\therefore\)实数\(m\)的取值范围是\([3,+∞)\).
分层练习
【A组---基础题】
1.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\(S_n=\)( )
A. \(\dfrac{n^2-1}{n}\) \(\qquad \qquad\) B. \(\dfrac{n(n+1)}{2}+2^n\) \(\qquad \qquad\) C. \(\dfrac{n(n+1)}{2}-\dfrac{1}{2^n}+1\)\(\qquad \qquad\) D. \(\dfrac{n}{2^n}-1\)
2.设 \(f(x)=\dfrac{1}{9^x+3}\),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\(\underline{\quad \quad}\).
3.数列\(\{a_n\}\)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\(\underline{\quad \quad}\).
4.已知等差数列\(\{a_n\}\)中,公差\(d>0\),\(a_1\),\(a_5\)为方程\(x^2-10x+9=0\)的两根.
(1)求数列\(\{a_n\}\)的通项公式;
(2)设\(b_n=a_n+ \dfrac{1}{2^n}\),求数列\(\{b_n\}\)的前\(n\)项和\(T_n\).
5.已知正项数列\(\left\{a_n\right\}\)满足 \(a_1=\sqrt{2}\), \(a_{n+1}^2-a_n^2=2(n+1)\).
(1)求数列\(\left\{a_n\right\}\)的通项公式;
(2)若数列\(\left\{b_n\right\}\)满足\(b_1=3\), \(b_n=3 b_{n-1}+3^n(n \geq 2)\),求数列 \(\left\{\dfrac{a_n^2}{b_n}\right\}\)的前\(n\)项和\(S_n\).
6.已知公差不为0的等差数列\(\{a_n\}\)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\(\{a_n\}\)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\(T_n\).
7.已知等差数列\(\{a_n\}\)的前\(n\)项和为\(S_n\),且\(S_2=8\),\(S_9=11a_4\).
(1)求\(a_n\);
(2)设数列\(\left\{\dfrac{1}{S_n}\right\}\)的前\(n\)项和为\(T_n\),求证:\(T_n<\dfrac{3}{4}\).
8.已知数列列\(\{a_n\}\)的前\(n\)项和为\(S_n\),且满足\(2S_n=3a_n-3\).
(1)证明数列\(\{a_n\}\)是等比数列;
(2)若数列\(\{b_n\}\)满足 \(b_n=\log _3 a_n\),记数列 \(\left\{\dfrac{b_n}{a_n}\right\}\)的前\(n\)项和为\(T_n\),证明: \(\dfrac{1}{3}<T_n<\dfrac{3}{4}\).
9.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\(S_n=f\left(\dfrac{1}{n}\right)+f\left(\dfrac{2}{n}\right)+\cdots+f\left(\dfrac{n-1}{n}\right)+f\left(\dfrac{n}{n}\right)\).
参考答案
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答案 \(C\)
解析 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
\(S_n=(1+2+3+\cdots+n)+\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\cdots+\dfrac{1}{2^n}\right)\)\(=\dfrac{n(n+1)}{2}+\dfrac{\dfrac{1}{2}\left(1-\dfrac{1}{2^n}\right)}{1-\dfrac{1}{2}}=\dfrac{n(n+1)}{2}-\dfrac{1}{2^n}+1\).
故选:\(C\). -
答案 \(\dfrac{4}{3}\)
解析 设\(a+b=1\),
则 \(f(a)+f(b)=\dfrac{1}{9^a+3}+\dfrac{1}{9^b+3}=\dfrac{1}{9^{1-b}+3}+\dfrac{1}{9^b+3}\)\(=\dfrac{9^b}{3\left(9^b+3\right)}+\dfrac{1}{9^b+3}=\dfrac{9^b+3}{3\left(9^b+3\right)}=\dfrac{1}{3}\) ,
所以\(f(-3)+f(4)= \dfrac{1}{3}\),\(f(-2)+f(3)= \dfrac{1}{3}\),\(f(-1)+f(2)= \dfrac{1}{3}\),\(f(0)+f(1)= \dfrac{1}{2}\),
\(f(-3)+f(-2)+⋯+f(0)+⋯+f(3)+f(4)=4× \dfrac{1}{3}=\dfrac{4}{3}\). -
答案 \(\dfrac{1}{6}\)
解析 \(a_n=\dfrac{1}{(2 n+1)(2 n+3)}=\dfrac{1}{2}\left(\dfrac{1}{2 n+1}-\dfrac{1}{2 n+3}\right)\),
可得其前\(n\)项和 \(S_n=\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\cdots+\dfrac{1}{2 n+1}-\dfrac{1}{2 n+3}\right)=\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{2 n+3}\right)\),
由\(\dfrac{1}{2 n+3}>0\),可得 \(S_n<\dfrac{1}{6}\),
\(S_n<M\)恒成立,可得 \(M \geq \dfrac{1}{6}\),即\(M\)的最小值为 \(\dfrac{1}{6}\). -
答案 (1)\(a_n=2n-1\);(2)\(T_n=n^2+1- \dfrac{1}{2^n}\) .
解析 等差数列\(\{a_n\}\)中,公差\(d>0\),\(a_1\),\(a_5\)为方程\(x^2-10x+9=0\)的两根.
所以\(a_1+a_5=10\),\(a_1 a_5=9\),解得\(a_1=1\),\(a_5=9\).
所以\(d=\dfrac{9-1}{5-1}=2\).
故\(a_n=a_1+2(n-1)=2n-1\).
(2)由于\(a_n=2n-1\),
所以\(b_n=2 n-1+\dfrac{1}{2^n}\),
所以\(T_n=(1+3+\cdots+2 n-1)+\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\cdots+\dfrac{1}{2^n}\right)\)\(=\dfrac{n(1+2 n-1)}{2}+\dfrac{\dfrac{1}{2}\left(1-\dfrac{1}{2^n}\right)}{1-\dfrac{1}{2}}=n^2+1-\dfrac{1}{2^n}\). -
答案 (1) \(a_n=\sqrt{n^2+n}\);(2) \(S_n=\dfrac{5}{4}-\dfrac{2 n+5}{4} \cdot\left(\dfrac{1}{3}\right)^n\).
解析 (1)由 \(a_1=\sqrt{2}\), \(a_{n+1}^2-a_n^2=2(n+1)\),
可得 \(a_n^2=a_1^2+\left(a_2^2-a_1^2\right)+\left(a_3^2-a_2^2\right)+\ldots+\left(a_n^2-a_{n-1}^2\right)\)
\(=2+4+6+\ldots+2 n=\dfrac{1}{2} n(2+2 n)=n^2+n\),
又 \(乙 a_n=0\),所以 \(a_n=\sqrt{n^2+n}\);
(2)\(b_1=3\), \(b_n=3 b_{n-1}+3^n(n \geq 2)\),
可得 \(\dfrac{b_n}{3^n}=\dfrac{b_{n-1}}{3^{n-1}}+1(n \geq 2)\),
则\(\left\{\dfrac{b_n}{3^n}\right\}\)是以 \(\dfrac{b_1}{3}=1\)为首项,公差为\(1\)的等差数列,
则\(\dfrac{b_n}{3^n}=n\),即为\(b_n=n\cdot 3^n\),
令\(c_n=\dfrac{a_n^2}{b_n}\),则 \(c_n=\dfrac{n+1}{3^n}\) ,
所以\(S_n=\dfrac{2}{3^1}+\dfrac{3}{3^2}+\cdots+\dfrac{n+1}{3^n}\) ,
\(\dfrac{1}{3} S_n=\dfrac{2}{3^2}+\dfrac{3}{3^3}+\cdots+\dfrac{n+1}{3^{n+1}}\),
两式相减可得\(\dfrac{2}{3} S_n=\dfrac{2}{3}+\dfrac{1}{3^2}+\cdots+\dfrac{1}{3^n}-\dfrac{n+1}{3^{n+1}}=\dfrac{1}{3}+\dfrac{\dfrac{1}{3}\left(1-\dfrac{1}{3^n}\right)}{1-\dfrac{1}{3}}-\dfrac{n+1}{3^{n+1}}\),
化为\(S_n=\dfrac{5}{4}-\dfrac{2 n+5}{4} \cdot\left(\dfrac{1}{3}\right)^n\). -
答案 (1)\(a_n=n\) (2) \(T_n=\dfrac{n^2+n+4}{2}-\dfrac{1}{2^{n-1}}\)
解析 (1)设公差为\(d\),且\(d≠0\)的等差数列\(\{a_n\}\)的前\(9\)项和\(S_9=45\),
且第\(2\)项、第\(4\)项、第\(8\)项成等比数列.
所以\(\left\{\begin{array}{c} S_9=45 \\ a_4^2=a_2 a_8 \end{array}\right.\),
整理得\(\left\{\begin{array}{c} 9 a_1+36 d=45 \\ \left(a_1+3 d\right)^2=\left(a_1+d\right)\left(a_1+7 d\right) \end{array}\right.\),解得\(a_1=d=1\),
故\(a_n=n\).
(2)由(1)得:数列\(\{b_n\}\)满足\(b_n=a_n+\left(\dfrac{1}{2}\right)^{n-1}=n+\left(\dfrac{1}{2}\right)^{n-1}\),
所以\(T_n=(1+2+\ldots+n)+\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+\ldots+\dfrac{1}{2^{n-1}}\right)=\dfrac{n^2+n+4}{2}-\dfrac{1}{2^{n-1}}\). -
答案 (1)\(a_n=2n+1\);(2) 略 .
解析 (1)设公差为\(d\),由\(S_2=8\),\(S_9=11a_4\),
得\(\left\{\begin{array}{l} 2 a_1+d=8 \\ 9 a_1+36 d=11 a_1+33 a \end{array}\right.\),解得\(a_1=3\),\(d=2\).
\(\therefore a_n=3+2(n-1)=2n+1\);
证明:(2)由(1),\(a_n=2n+1\),则有\(S_n=\dfrac{n}{2}(3+2 n+1)=n^2+2 n\).
则\(\dfrac{1}{S_n}=\dfrac{1}{n(n+2)}=\dfrac{1}{2}\left(\dfrac{1}{n}-\dfrac{1}{n+2}\right)\).
\(\therefore T_n=\dfrac{1}{2}\left[\left(1-\dfrac{1}{3}\right)+\left(\dfrac{1}{2}-\dfrac{1}{4}\right)+\left(\dfrac{1}{3}-\dfrac{1}{5}\right)+\ldots+\left(\dfrac{1}{n-1}-\dfrac{1}{n+1}\right)+\left(\dfrac{1}{n}-\dfrac{1}{n+2}\right)\right]\)
\(=\dfrac{1}{2}\left(1+\dfrac{1}{2}-\dfrac{1}{n+1}-\dfrac{1}{n+2}\right)<\dfrac{3}{4}\). -
答案 (1) 略;(2) 略 .
解析 (1)因为\(2S_n=3a_n-3\),所以 \(2 S_{n-1}=3 a_{n-1}-3\),
两式相减得,\(2a_n=3a_n-3a_{n-1} (n≥2)\),即\(a_n=3a_{n-1} (n≥2)\),
在\(2S_n=3a_n-3\)中,令\(n=1\),则\(2a_1=2S_1=3a_1-3\),解得\(a_1=3≠0\),
故数列\(\{a_n\}\)是以\(3\)为首项,\(3\)为公比的等比数列.
(2)由(1)可知\(a_n=3^n\).
所以 \(b_n=\log _3 a_n=\log _3 3^n=n\),所以 \(\dfrac{b_n}{a_n}=\dfrac{n}{3^n}\).
所以 \(T_n=\dfrac{1}{3^1}+\dfrac{2}{3^2}+\dfrac{3}{3^3}+\cdots \cdots+\dfrac{n-1}{3^{n-1}}+\dfrac{n}{3^n}\),
\(\dfrac{1}{3} T_n=\dfrac{1}{3^2}+\dfrac{2}{3^3}+\dfrac{3}{3^4}+\cdots \cdots+\dfrac{n-1}{3^n}+\dfrac{n}{3^{n+1}}\) ,
两式相减得, \(\dfrac{2}{3} T_n=\dfrac{1}{3^1}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+\cdots \cdots+\dfrac{1}{3^n}-\dfrac{n}{3^{n+1}}\)\(=\dfrac{\dfrac{1}{3}\left[1-\left(\dfrac{1}{3}\right)^n\right]}{1-\dfrac{1}{3}}-\dfrac{n}{3^{n+1}}=\dfrac{3^{n+1}-2 n-3}{2 \cdot 3^{n+1}}\),
所以 \(T_n=\dfrac{3^{n+1}-2 n-3}{4 \cdot 3^n}=\dfrac{3}{4}-\dfrac{2 n+3}{4} \cdot \dfrac{1}{3^n}<\dfrac{3}{4}\),
当\(n≥2\)时, \(T_n-T_{n-1}=\dfrac{b_n}{a_n}=\dfrac{n}{3^n}>0\),
故数列\(\left\{T_n\right\}\)为递增数列, \(T_n>T_1=\dfrac{1}{3}\).
综上所述,\(\dfrac{1}{3}<T_n<\dfrac{3}{4}\). -
答案 (1) \(\dfrac{1}{2}\);(2) \(S_n=\dfrac{n+3-2 \sqrt{2}}{2}\) .
解析 (1)证: \(\because \overrightarrow{O P}=\dfrac{1}{2}\left(\overrightarrow{O P_1}+\overrightarrow{O P_2}\right)\),
\(\therefore P\)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\(x_1+x_2=1\),\(f (x_1)+f (x_2)=y_1+y_2=1\), \(f(1)=2-\sqrt{2}\),
由 \(S_n=f\left(\dfrac{1}{n}\right)+f\left(\dfrac{2}{n}\right)+\cdots+f\left(\dfrac{n-1}{n}\right)+f\left(\dfrac{n}{n}\right)\),
\(S_n=f\left(\dfrac{n}{n}\right)+f\left(\dfrac{n-1}{n}\right)+\cdots+f\left(\dfrac{2}{n}\right)+f\left(\dfrac{1}{n}\right)\),
相加得
\(2 S_n=f(1)+\left[f\left(\dfrac{1}{n}\right)+f\left(\dfrac{n-1}{n}\right)\right]+\left[f\left(\dfrac{2}{n}\right)+f\left(\dfrac{n-2}{n}\right)\right]+\)\(\cdots+\left[f\left(\dfrac{n-1}{n}\right)+f\left(\dfrac{1}{n}\right)\right]+f(1)=2 f(1)+n-1=n+3-2 \sqrt{2}\),
\(\therefore S_n=\dfrac{n+3-2 \sqrt{2}}{2}\).
【B组---提高题 】
1.求 \(\sin ^2 1^{\circ}+\sin ^2 2^{\circ}+\sin ^2 3^{\circ}+\cdots+\sin ^2 88^{\circ}+\sin ^2 89^{\circ}\)的值.
2.已知数列\(\{a_n\}\)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\(S_n\)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\(\{a_n\}\)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\(n∈N^*\),\(T_n≤nS_{n+1}-m-2\)恒成立,求实数\(m\)的取值范围.
3.已知数列\(\{a_n\}\)满足\(a_n≠0\),\(a_1= \dfrac{1}{3}\),\(a_{n-1}-a_n=2a_n a_{n-1} (n≥2 ,n∈N^*)\).
(1)求证:\(\left\{\dfrac{1}{a_n}\right\}\)是等差数列;\(\qquad \qquad\) (2)证明:\(a_1^2+a_2^2+⋯+a_n^2< \dfrac{1}{4}\).
参考答案
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答案 \(44.5\)
解析 设 \(S=\sin ^2 1^{\circ}+\sin ^2 2^{\circ}+\sin ^2 3^{\circ}+\cdots+\sin ^2 88^{\circ}+\sin ^2 89^{\circ}\)………… 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
\(2 S=\left(\sin ^2 1^{\circ}+\sin ^2 89^{\circ}\right)+\left(\sin ^2 2^{\circ}+\sin ^2 88^{\circ}\right)+\cdots+\left(\sin ^2 89^{\circ}+\sin ^2 1^{\circ}\right)\)
\(=\left(\sin ^2 1^{\circ}+\cos ^2 1^{\circ}\right)+\left(\sin ^2 2^{\circ}+\cos ^2 2^{\circ}\right)+\cdots+\left(\sin ^2 89^{\circ}+\cos ^2 89^{\circ}\right)=89\)
\(\therefore S=44.5\). -
答案 (1) 证明路,\(a_n=(n+1)^2\),\(b_n=2^{n-1}\);(2) \(m≤-1\) .
解析 (1)证明: \(\because(n+1) a_{n+1}-(n+2) a_n=(n+1)(n+2)\),
\(\therefore \dfrac{a_{n+1}}{n+2}-\dfrac{a_n}{n+1}=1\).\(\therefore\)数列\(\left\{\dfrac{a_n}{n+1}\right\}\)为等差数列,
又\(a_1=4\), \(\therefore \dfrac{a_ 1}{2}=2\),公差为\(1\).
\(\therefore \dfrac{a_n}{n+1}=2+(n-1) \times 1=n+1\),则\(a_n=(n+1)^2\);
\(\because S_n=2b_n-1\),①
令\(n=1\),得\(b_1=1\),
\(\therefore S_{n+1}=2b_{n+1}-1\),②
由②-①得:\(S_{n+1}-S_n=2b_{n+1}-2b_n=b_{n+1}\).
\(\therefore \dfrac{b_{n+1}}{b_n}=2\).
\(\therefore \{b_n\}\)为等比数列,且\(b_1=1\),\(q=2\).
\(\therefore b_n=2^{n-1}\);
(2) \(c_n=\left(\sqrt{a_n}-1\right) b_{n+1}=n 2^n\).
\(\therefore T_n=1 \times 2+2 \times 2^2+3 \times 2^3+\cdots+(n-1) \times 2^{n-1}+n \times 2^n\),①
\(\therefore 2T_n=1×2^2+2×2^3+⋯+{n-1}×2^n+n×2^{n+1}\),②
由①-②得:
\(-T_n=2+2^2+2^3+\cdots+2^n-n \times 2^{n+1}=\dfrac{2\left(1-2^n\right)}{1-2}-n \times 2^{n+1}\).
\(\therefore T_n=(n-1)×2^{n+1}+2\).
而\(S_n=2b_n-1=2^n-1\), \(\therefore S_{n+1}=2^{n+1}-1\).
\(\because T_n≤nS_{n+1}-m-2\)恒成立,
\(\therefore m≤nS_{n+1}-2-T_n=2^{n+1}-n-4\).
令\(f(n)=2^{n+1}-n-4\),
\(\therefore f(n+1)-f(n)=2^{n+1}-1>0\).
\(\therefore f(n)\)递增,则 \(f(n)_{\min }=f(1)=-1\).
故\(m≤-1\). -
证明 (1)\(\because a_{n-1}-a_n=2a_n a_{n-1} (n≥2 ,n∈N^*)\)
\(\therefore \dfrac{1}{a_n}-\dfrac{1}{a_{n-1}}=2(n \geq 2)\),
\(\therefore\left\{\dfrac{1}{a_n}\right\}\)是以\(3\)为首项,\(2\)为公差的等差数列.
(2)由(1)知: \(\dfrac{1}{a_n}=3+(n-1) \cdot 2=2 n+1\),
\(\therefore a_n=\dfrac{1}{2 n+1}\) ,
\(\therefore a_n^2=\dfrac{1}{(2 n+1)^2}<\dfrac{1}{4 n^2+4 n}=\dfrac{1}{4 n(n+1)}=\dfrac{1}{4}\left(\dfrac{1}{n}-\dfrac{1}{n+1}\right)\),
\(\therefore a_1^2+a_2^2+\cdots+a_n^2<\dfrac{1}{4}\left(\dfrac{1}{1}-\dfrac{1}{2}\right)+\dfrac{1}{4}\left(\dfrac{1}{2}-\dfrac{1}{3}\right)+\cdots+\dfrac{1}{4}\left(\dfrac{1}{n}-\dfrac{1}{n+1}\right)\)
\(<\dfrac{1}{4}\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\cdots+\dfrac{1}{n}-\dfrac{1}{n+1}\right)=\dfrac{1}{4}\left(1-\dfrac{1}{n+1}\right)<\dfrac{1}{4}\).
【C组---拓展题 】
1.设\(f(x)=(x-1)^3+1\),则\(f(-4)+⋯+f(0)+⋯+f(5)+f(6)\)的值为\(\underline{\quad \quad}\).
2.已知正项数列\(\{a_n\}\)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\(\{b_n\}\)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\(\{a_n\}\)满足\(a_1=1\), \(\sqrt{a_n}-\sqrt{a_{n+1}}=\sqrt{a_n \cdot a_{n+1}}\left(n \in N^*\right)\).
(1)求证:数列\(\left\{\sqrt{\dfrac{1}{a_n}}\right\}\)为等差数列,并求\(a_n\);
(2)设 \(b_n=\sqrt{1+2 a_{n+1}}\),数列\(\{b_n\}\)的前\(n\)项和为\(S_n\),求证:\(S_n<n+1-\dfrac{1}{n+1}\).
参考答案
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答案 \(11\)
解析 用倒序相加法:
令\(f(-4)+f(-3)+⋯+f(0)+⋯+f(5)+f(6)=S\) ①
则也有\(f(6)+f(5)+⋯+f(0)+⋯+f(-3)+f(-4)=S\) ②
由\(f(x)+f(2-x)=(x-1)^3+1+(x-1)^3+1=2\),
可得:\(f(-4)+f(6)=f(-3)+f(5)=⋯=2\),
于是由①②两式相加得\(2S=11×2\),
所以\(S=11\) . -
答案 \(9\)
解析 \(\because 2S_n=a_n^2+a_n\),
\(\therefore\)当\(n≥2\)时,\(2a_n=2(S_n-S_{n-1} )=(a_n^2+a_n )-(a_{n-1}^2+a_{n-1})\),
整理得:\((a_n-a_{n-1})( a_n+a_{n-1})=a_n+a_{n-1}\),
又\(\because\)数列\(\{a_n\}\)的每项均为正数,\(\therefore a_n-a_{n-1}=1\),
又\(\because 2a_1=a_1^2+a_1\),即\(a_1=1\),
\(\therefore\)数列\(\{a_n\}\)是首项、公差均为\(1\)的等差数列, \(\therefore a_n=n\),
\(\therefore b_n=\dfrac{1}{a_n \sqrt{a_{n+1}}+a_{n+1} \sqrt{a_n}}=\dfrac{1}{\sqrt{n(n+1)}} \cdot \dfrac{1}{\sqrt{n}+\sqrt{n+1}}\)\(=\dfrac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n(n+1)}}=\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\),
\(\therefore\) 数列\(\{b_n\}\)的前\(n\)项和为 \(T_n=1-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+\cdots+\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}=1-\dfrac{1}{\sqrt{n+1}}\),
要使得\(T_n\)为有理数,只需\(\dfrac{1}{\sqrt{n+1}}\)为有理数即可,即\(n+1=t^2\),
\(\because 1≤n≤100\),\(\therefore t=3、8、15、24、35、48、63、80、99\),
即在\(T_1\),\(T_2\),\(T_3\),\(…\),\(T_{100}\)中有理数的个数为\(9\)个,
故答案为:\(9\). -
答案 (1) 证明略,\(a_n=\dfrac{1}{n ^2}\) ;(2) 略.
解析 (1)解:由\(\sqrt{a_n}-\sqrt{a_{n+1}}=\sqrt{a_n \cdot a_{n+1}}\),得\(\dfrac{1}{\sqrt{a_{n+1}}}-\dfrac{1}{\sqrt{a_n}}=1\),
所以数列\(\left\{\sqrt{\dfrac{1}{a_n}}\right\}\)是以\(1\)为首项\(1\)为公差的等差数列,
即\(\dfrac{1}{\sqrt{a_n}}=1+(n-1)=n\),化简得 \(a_n=\dfrac{1}{n^2}\).
(2)证明:因为 \(\dfrac{1}{n+1}-\dfrac{1}{n}<0\),
所以\(\dfrac{2}{n+1}\left(\dfrac{1}{n+1}-\dfrac{1}{n}\right)<0<\dfrac{1}{n^2(n+1)^2}\),
可得:\(\dfrac{2}{(n+1)^2}<\dfrac{2}{n(n+1)}+\dfrac{1}{n^2(n+1)^2}\),
即\(1+\dfrac{2}{(n+1)^2}<1+\dfrac{2}{n(n+1)}+\dfrac{1}{n^2(n+1)^2}\),
所以\(\sqrt{1+\dfrac{2}{(n+1)^2}}<1+\dfrac{1}{n(n+1)}\),
因为\(b_n=\sqrt{1+2 a_{n+1}}=\sqrt{1+\dfrac{2}{(n+1)^2}}<1+\dfrac{1}{n(n+1)}\),
所以\(S_n=b_1+b_2+\cdots+b_n\)\(<n+\dfrac{1}{1 \times 2}+\dfrac{1}{2 \times 3}+\cdots+\dfrac{1}{n(n+1)}=n+1-\dfrac{1}{n+1}\).
所以\(S_n<n+1-\dfrac{1}{n+1}\).