4.4 数学归纳法
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基础知识
数学归纳法的概念
一般地,证明一个与正整数\(n\)有关的命题,可按下列步骤进行:
(1)(归纳奠基)证明当\(n=n_0\)\((n_0∈N^*)\)时命题成立;
(2)(归纳递推)以“当\(n=k(k∈N^* ,k≥n_0)\)时命题成立”为条件,推出“当\(n=k+1\)时命题也成立”;
只要完成这两个步骤,就可以断定命题对从\(n_0\)开始的所有正整数\(n\)都成立,这种证明方法称为数学归纳法.
解释
1.数学概念法类似“多米诺骨牌”,满足两个条件:①第一个骨牌可倒下;②任一个骨牌倒下时均可令下个骨牌倒下;这样所有骨牌均倒下了!故用数学归纳法证明,两个步骤缺一不可.
2.第一步归纳奠基中的\(n_0\)不一定是\(1\);第二步中当证明从\(n=k\)到\(n=k+1\)时,所证明的式子不一定只增加一项.
3.在运用数学归纳法证明在证明第二步中,强调两个“凑”,一是“凑”假设,在\(n=k+1\)时的式子中凑出\(n=k\)的式子(确定两个式子的“差项”;二是“凑”结论,明确\(n=k+1\)时要证明的目标,在这个过程中常用到比较法、分析法等,不等式证明中还会用到放缩法);
4.要注意“观察---归纳—猜想---证明”的思维模式和由特殊到一般的数学思想.
数学归纳法的运用
数学归纳法证明的对象是与正整数\(n\)有关的命题,比如:与正整数\(n\)有关的等式或不等式的证明,求数列的通项公式,与数列有关的不等关系证明,整除问题,函数不等式等.
用数学归纳法证明\(1+2+\cdots+n=\dfrac{n(n+1)}{2}\left(n \in N^*\right)\).
证明 当\(n=1\)时,左边\(=1\),右边 \(=\dfrac{1 \times 2}{2}=1\),故等式成立;
假设\(n=k(k≥1,k∈N^*)\)时,等式成立,即 \(1+2+\cdots+k=\dfrac{k(k+1)}{2}\);
当\(n=k+1\)时,左边 \(=1+2+\cdots+k+(k+1)=\dfrac{k(k+1)}{2}+(k+1)=\dfrac{(k+1)(k+2)}{2}\),
右边 \(=\dfrac{(k+1)(k+2)}{2}\), \(\therefore\)左边=右边,故等式成立.
综上,\(1+2+\cdots+n=\dfrac{n(n+1)}{2}\left(n \in N^*\right)\)成立.
基本方法
【题型1】 对数学归纳法的理解
【典题1】 用数学归纳法证明“\(2^n>n+2\)对于\(n≥n_0\)的正整数\(n\)都成立”时,第一步证明中的起始值\(n_0\)应取\(\underline{\quad \quad}\).
解析 根据数学归纳法的步骤,首先要验证当n取第一个值时命题成立;
结合本题,要验证\(n=1\)时,左边\(=2^1=2\),右边\(=1+2=3\),\(2^n>n+2\)不成立,
\(n=2\)时,左边\(=2^2=4\),右边\(=2+2=4\),\(2^n>n+2\)不成立,
\(n=3\)时,左边\(=2^3=8\),右边\(=3+2=5\),\(2^n>n+2\)成立,
\(n=4\)时,左边\(=2^4=16\),右边\(=4+2=6\),\(2^n>n+2\)成立,
…
因为\(n>2\)成立,所以\(2^n>n+2\)恒成立.
故\(n_0=3\).
点拨 数学归纳法第一步中的\(n_0\)不一定是\(1\),一般是满足题意的最小的正整数.
【典题2】 用数学归纳法证明命题“当\(n\)是正奇数时,\(x^n+y^n\)能被\(x+y\)整除”,在第二步时,正确的证法是 ( )
A.假设\(n=k\)\((k∈N^*)\),证明\(n=k+1\)命题成立
B.假设\(n=k\)\((k\)是正奇数\()\),证明\(n=k+1\)命题成立
C.假设\(n=2k+1(k∈N^*)\),证明\(n=k+1\)命题成立
D.假设\(n=k\)(k是正奇数),证明\(n=k+2\)命题成立
解析 \(A\)、\(B\)、\(C\)中,\(k+1\)不一定表示奇数,只有\(D\)中\(k\)为奇数,\(k+2\)为奇数.
故答案:\(D\).
点拨 注意第二步中不一定是\(n=k+1\),要注意题目对\(n\)的要求.
【巩固练习】
1.已知数列\(\{a_n\}\)中,\(a_1=1\), \(a_{n+1}=1+\dfrac{a_n}{1+a_n}\left(n \in \boldsymbol{N}^*\right)\),用数学归纳法证明:\(a_n<a_{n+1}\),在验证\(n=1\)成立时,不等式右边计算所得结果是( )
A.\(\dfrac{1}{2}\) \(\qquad \qquad \qquad \qquad\) B.\(1\) \(\qquad \qquad \qquad \qquad\) C. \(\dfrac{3}{2}\) \(\qquad \qquad \qquad \qquad\) D.\(2\)
2.用数学归纳法证明\(2^n≥n^2 (n≥4)\)时,第二步应假设( )
A.\(n=k≥2\)时,\(2^k≥k^2\) \(\qquad \qquad \qquad \qquad\) B.\(n=k≥3\)时,\(2^k≥k^2\)
C.\(n=k≥4\)时,\(2^k≥k^2\) \(\qquad \qquad \qquad \qquad\) D.\(n=k≥5\)时,\(2^k≥k^2\)
3.用数学归纳法证明不等式\(\dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{4} +⋯+ \dfrac{1}{2^{n-1}} > \dfrac{n}{2}-1\)\((n∈N^*,n>1)\)时,以下说法正确的是 ( )
A.第一步应该验证当\(n=1\)时不等式成立
B.从“\(n=k\)到\(n=k+1\)”左边需要增加的代数式是\(\dfrac{1}{2 ^k}\)
C.从“\(n=k\)到\(n=k+1\)”左边需要增加\(\left(2^{k-1}-1\right)\)项
D.从“\(n=k\)到\(n=k+1\)”左边需要增加 \(2^{k-1}\)项
参考答案
-
答案 \(C\)
解析 由题设可知:当\(n=1\)时,不等式右边为\(a_2=1+\dfrac{a_1}{1+a_1}=1+\dfrac{1}{2}=\dfrac{3}{2}\),故选:\(C\). -
答案 \(C\)
解析 根据证明的结论,\(n≥4\),
故第二步的假设应写成:假设\(n=k\),\(n≥4\),\(k∈N^*\)时命题正确,即\(2^k≥k^2\)正确.
故选:\(C\). -
答案 \(D\)
解析 由于\(n∈N^*\),\(n>1\),
所以第一步应该是验证当\(n=2\)时不等式成立,
从“\(n=k\)到\(n=k+1\)”左边需要增加的代数式是 \(\dfrac{1}{2^{k-1}+1}+\dfrac{1}{2^{k-1}+2}+\cdots+\dfrac{1}{2^k}\),共 \(2^{k-1}\)项.
故选:\(D\).
【题型2】 数学归纳法的运用
【典题1】 用数学归纳法证明 \(\dfrac{1}{1 \times 2}+\dfrac{1}{3 \times 4}+\cdots+\dfrac{1}{(2 n-1) \times 2 n}=\dfrac{1}{n+1}+\dfrac{1}{n+2}+\cdots+\dfrac{1}{n+n}\).
证明 (1)当\(n=1\)时,左边 \(=\dfrac{1}{1 \times 2}=\dfrac{1}{2}\) ,右边\(= \dfrac{1}{2}\),等式成立.
(2)假设当\(n=k\)时,等式成立,
即\(\dfrac{1}{1 \times 2}+\dfrac{1}{3 \times 4}+\cdots+\dfrac{1}{(2 k-1) \times 2 k}=\dfrac{1}{k+1}+\dfrac{1}{k+2}+\cdots+\dfrac{1}{2 k}\)
则当\(n=k+1\)时,
\(\dfrac{1}{1 \times 2}+\dfrac{1}{3 \times 4}+\cdots+\dfrac{1}{(2 k-1) \times 2 k}+\dfrac{1}{(2 k+1) \times(2 k+2)}\)
\(=\dfrac{1}{k+1}+\dfrac{1}{k+2}+\cdots+\dfrac{1}{2 k}+\dfrac{1}{(2 k+1) \times(2 k+2)}\)
\(=\dfrac{1}{k+2}+\dfrac{1}{k+3}+\cdots+\dfrac{1}{2 k}+\left(\dfrac{1}{2 k+1}-\dfrac{1}{2 k+2}\right)+\dfrac{1}{k+1}\)
\(=\dfrac{1}{k+2}+\dfrac{1}{k+3}+\cdots+\dfrac{1}{2 k}+\dfrac{1}{2 k+1}+\dfrac{1}{2 k+2}\)
\(=\dfrac{1}{(k+1)+1}+\dfrac{1}{(k+1)+2}+\cdots+\dfrac{1}{(k+1)+k}+\dfrac{1}{(k+1)(k+1)}\)
即当\(n=k+1\)时,等式成立.
根据(1)、(2)可知,对一切\(n∈N^*\),等式成立.
【典题2】 试用数学归纳法证明\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\cdots+\dfrac{1}{(n+1)^2}>\dfrac{1}{2}-\dfrac{1}{n+2}\).
证明 (1)当\(n=1\)时,左边\(= \dfrac{1}{4}\),右边 \(=\dfrac{1}{6}\),不等式成立
(2)假设当\(n=k\)时,原式成立,即 \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\cdots+\dfrac{1}{(k+1)^2}>\dfrac{1}{2}-\dfrac{1}{k+2}\),
当\(n=k+1\)时, \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\cdots+\dfrac{1}{(k+1)^2}+\dfrac{1}{(k+2)^2}>\dfrac{1}{2}-\dfrac{1}{k+2}+\dfrac{1}{(k+2)^2}\)
\(\because-\dfrac{1}{k+2}+\dfrac{1}{(k+2)^2}+\dfrac{1}{k+3}=\dfrac{1}{(k+2)^2(k+3)}>0\),
\(\therefore-\dfrac{1}{k+2}+\dfrac{1}{(k+2)^2}>-\dfrac{1}{k+3}\),
\(\therefore \dfrac{1}{2}-\dfrac{1}{k+2}+\dfrac{1}{(k+2)^2}>\dfrac{1}{2}-\dfrac{1}{k+3}\),
即\(n=k+1\)时结论成立.
根据(1)和(2)可知不等式对任意正整数\(n\)都成立
\(\therefore \dfrac{1}{2^2}+\dfrac{1}{3^2}+\cdots+\dfrac{1}{(n+1)^2}>\dfrac{1}{2}-\dfrac{1}{n+2}\).
【典题3】 已知数列\(\{a_n\}\)的前\(n\)项和\(S_n=2n-a_n\).
(1)计算\(a_1,a_2,a_3,a_4\),并猜\(\{a_n\}\)的通项公式;\(\qquad \qquad\)(2)用数学归纳法证明(1)中的猜想.
解析 (1)根据题意,\(S_n=2n-a_n\).
当\(n=1\)时,\(a_1=S_1=2-a_1\),\(\therefore a_1=1\);
当\(n=2\)时,\(a_1+a_2=S_2=2×2-a_2\), \(\therefore a_2= \dfrac{3}{2}\) ;
当\(n=3\)时,\(a_1+a_2+a_3=S_3=2×3-a_3\), \(\therefore a_3=\dfrac{7}{4}\);
当\(n=4\)时,\(a_1+a_2+a_3+a_4=S_4=2×4-a_4\), \(\therefore a_4=\dfrac{15}{8}\).
由此猜想 \(a_n=\dfrac{2^n-1}{2^{n-1}}\);
(2)证明:①当\(n=1\)时,\(a_1=1\),猜想成立.
②假设\(n=k\)\((k≥1\)且\(k∈N^*)\)时,猜想立,即 \(a_k=\dfrac{2^k-1}{2^{k-1}}\),
那么\(n=k+1\)时,
\(a_{k+1}=S_{k+1}-S_k=2(k+1)-a_{k+1}-2 k+a_k=2+a_k-a_{k+1}\),
\(\therefore a_{k+1}=\dfrac{2+a_k}{2}=\dfrac{2+\dfrac{2^k-1}{2^{k-1}}}{2}=\dfrac{2^{k+1}-1}{2^k}\) .
\(\therefore\)当\(n=k+1\)时,猜想成立.
由①②知猜想\(a_n=\dfrac{2^n-1}{2^{n-1}}\left(n \in N^*\right)\)成立.
【巩固练习】
1.用数学归纳法证明:\(1×4+2×7+⋯+n(3n+1)=n(n+1)^2\)
2.若\(n\)属于自然数,\(n≥3\),证明:\(2^n>2n+1\).
3.用数学归纳法证明: \(1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\cdots+\dfrac{1}{2^n-1} \leq n(n \geq 1)\).
4.已知数列\(\{a_n\}\)满足\(a_1=1\), \(a_{n+1}=\dfrac{a_n}{1+a_n}\).
(1)计算\(a_2,a_3,a_4\);\(\qquad \qquad\)(2)猜测\(a_n\)的表达式,并用数学归纳法证明.
5.已知数列\(\{a_n\}\)的前\(n\)项和分别为\(S_n\),且\(a_n>0\),\(6S_n=a_n^2+3a_n\).
(1)求\(a_1,a_2,a_3\);\(\qquad \qquad\)(2)猜想\(a_n\)的表达式,并用数学归纳法证明.
参考答案
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证明 (1)当\(n=1\)时,左边\(=1×4=4\),右边\(=1×2^2=4\),所以等式成立(2)
假设当\(n=k\)时命题成立,即\(1×4+2×7+⋯+k(3k+1)=k(k+1)^2\),
那么,当\(n=k+1\)时,\(1×4+2×7+⋯+k(3k+1)+(k+1)(3k+4)\)\(=k(k+1)^2+(k+1)(3k+4)=(k+1) [(k+1)+1]^2\)
即\(n=k+1\)时,命题成立
由(1)(2)知等式对任意的\(n∈N^*\)均成立. -
证明 ①\(n=3\)时,\(8>7\)成立;
②假设\(n=k\)时不等式成立,即\(2^k>2k+1\);
则当\(n=k+1\)时,左边\(=2^{k+1}>4k+2>2k+3=2(k+1)+1\),成立
综上所述,\(2^n>2n+1\). -
证明 (1)当\(n=1\)时,左边\(=1\),右边\(=1\),命题成立.
(2)假设当\(n=k\)时, \(1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\cdots+\dfrac{1}{2^k-1} \leq k\)成立
当\(n=k+1\)时,左边 \(=1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\cdots+\dfrac{1}{2^k-1}+\dfrac{1}{2^k}+\cdots+\dfrac{1}{2^{k+1}-1}\)
\(\leq k+\dfrac{1}{2^k}+\cdots+\dfrac{1}{2^{k+1}-1} \leq k+\dfrac{1}{2^k}+\cdots+\dfrac{1}{2^k}=k+1\),
当\(n=k+1\)时命题成立.
由(1)(2)可得,对于任意\(n≥1\),\(n∈N^*\)都成立. -
答案 (1) \(a_2= \dfrac{1}{2}\),\(a_3= \dfrac{1}{3}\),\(a_4= \dfrac{1}{4}\);(2) \(a_n= \dfrac{1}{n}\),证明略.
解析 (1)解:由\(a_{n+1}=\dfrac{a_n}{1+a_n}\)及\(a_1=1\),得 \(a_2=\dfrac{a_1}{1+a_1}=\dfrac{1}{2}\),
进而 \(a_3=\dfrac{a_2}{1+a_2}=\dfrac{1}{3}\), \(a_4=\dfrac{a_3}{1+a_3}=\dfrac{1}{4}\),
(2)证明:猜想\(a_n= \dfrac{1}{n}\),再用数学归纳法证明之.
当\(n=1\)时, \(a_1=\dfrac{1}{1}=1\),而已知\(a_1=1\),所以\(n=1\)时,猜想正确.
假设当\(n=k\)时,猜想正确,即\(a_k= \dfrac{1}{k}\),
则\(n=k+1\)时, \(a_{k+1}=\dfrac{a_k}{1+a_k}=\dfrac{\dfrac{1}{k}}{1+\dfrac{1}{k}}=\dfrac{1}{k+1}\).
所以当\(n=k+1\)时,猜想也成立.
综上所述可知,对一切\(n∈N\),猜想\(a_n= \dfrac{1}{n}\)都正确. -
答案 (1) \(a_1=3\),\(a_2=6\),\(a_3=9\);(2) \(a_n=3n\),证明略 .
解析 (1)\(\because 6S_n=a_n^2+3a_n\),
\(\therefore\)当\(n=1\)时,\(6S_1=6a_1=a_1^2+3a_1\),
\(\because a_n>0\), \(\therefore a_1=3\),
当\(n=2\)时,\(6S_2=6(a_1+a_2)=a_2^2+3a_2\),
\(\because a_n>0\),\(\therefore a_2=6\),
当\(n=3\)时,\(6S_3=6(a_1+a_2+a_3)=a_3^2+3a_3\),
\(\because a_n>0\),\(\therefore a_3=9\),
故\(a_1=3\),\(a_2=6\),\(a_3=9\).
(2)猜想\(a_n=3n\),
证明:①当\(n=1\)时,左边\(a_1=3\),右边\(=3×1=3\),符合要求.
②假设当\(n=k\)时,\(a_k=3k\)
当\(n=k+1\)时, \(a_{k+1}=S_{k+1}-S_k=\dfrac{1}{6}\left(a_{k+1}^2+3 a_{k+1}-a_k^2-3 a_k\right)\)
即 \(a_{k+1}^2-3 a_{k+1}-9 k(k+1)=0\),
\(\because a_n>0\),即 \(a_{k+1}=3(k+1)\),
\(\therefore\) 当\(n=k+1\)时,也成立.
根据①②可知,\(a_n=3n\),即得证.
分层练习
【A组---基础题】
1.用数学归纳法证明\(1+2+3+⋯+(2n+1)={n+1}(2n+1)\)时,第一步当\(n=1\)时,左边的代数式是( )
A.\(1\) \(\qquad \qquad \qquad\) B.\(1+2\) \(\qquad \qquad \qquad\) C.\(1+2+3\) \(\qquad \qquad \qquad\) D.\(1+2+3+4+5\)
2.用数学归纳法证明“ \(1+\dfrac{1}{2}+\dfrac{1}{3}+\cdots+\dfrac{1}{2^n-1}<n\left(n \in \boldsymbol{N}^*, n>1\right)\)”时,由\(n=k(k>1)\)不等式成立,推证\(n=k+1\)时,左边应增加的项数是 ( )
A.\(2^{k-1}\) \(\qquad \qquad \qquad \qquad\) B.\(2^k-1\) \(\qquad \qquad \qquad \qquad\) C.\(2^k\) \(\qquad \qquad \qquad \qquad\) D.\(2^k+1\)
3.用数学归纳法证明某不等式时,其左边\(y=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+\cdots+\dfrac{1}{2 n-1}-\dfrac{1}{2 n}\),则从“\(n=k\)到\(n=k+1\)”应将左边加上\(\underline{\quad \quad}\).
4.用数学归纳法证明:
\(1×2×3+2×3×4+3×4×5+⋯+n(n+1)(n+2)= \dfrac{1}{4} n(n+1)(n+2)(n+3)\).
5.当\(n≥2\),\(n∈N^*\)时,求证:\(1+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+\cdots+\dfrac{1}{\sqrt{n}}>\sqrt{n}\).
6.已知数列\(\{a_n\}\)的前\(n\)项和为\(S_n\),且满足\(S_n=2a_n-n\).
(1)求\(a_1,a_2,a_3\),并猜想数列\(\{a_n\}\)的通项公式;
(2)用数学归纳法证明你的猜想.
7.证明:对一切\(n∈N^*\),都有\(2^{n+2}×3^n+5n-4\)能被\(25\)整除.
8.已知数列\(\{a_n\}\)满足 \(a_1=\dfrac{2}{5}\),\(a_{n+1} a_n+2a_{n+1}=2a_n\),\((n∈N^*)\).
(1)计算\(a_2,a_3,a_4\)的值;
(2)猜想数列\(\{a_n\}\)的通项公式,并用数学归纳法证明.
9.已知前三个式子分别为: \(1+\dfrac{1}{2^2}<\dfrac{3}{2}\) , \(1+\dfrac{1}{2^2}+\dfrac{1}{3^2}<\dfrac{5}{3}\), \(1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}<\dfrac{7}{4}\),….
(1)照此规律,写出第\(n\)个不等式;
(2)用数学归纳法和放缩法两种方法证明.
参考答案
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答案 \(C\)
解析 \(1+2+3+⋯+(2n+1)={n+1}(2n+1)\),
所以当\(n=1\)时,左边的代数式是\(1+2+3\).
故选:\(C\). -
答案 \(C\)
解析 增加的项数为 \(\left(2^{k+1}-1\right)-\left(2^k-1\right)=2^{k+1}-2^k=2^k\).答案:\(C\). -
答案 \(\dfrac{1}{2 k+1}-\dfrac{1}{2 k+2}\)
解析 当\(n=k\)时, \(f(k)=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+\ldots+\dfrac{1}{2 k-1}-\dfrac{1}{2 k}\),
当\(n=k+1\)时, \(f(k+1)=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+\ldots+\dfrac{1}{2 k-1}-\dfrac{1}{2 k}+\dfrac{1}{2 k+1}-\dfrac{1}{2 k+2}\),
所以\(f(k+1)-f(k)=\dfrac{1}{2 k+1}-\dfrac{1}{2 k+2}\).
故答案为: \(\dfrac{1}{2 k+1}-\dfrac{1}{2 k+2}\). -
证明 当\(n=1\)时,\(1×2×3= \dfrac{1}{4} ×1×2×3×4=6\),等式成立.
假设当\(n=k\)时等式成立,
即\(1×2×3+2×3×4+3×4×5+⋯+k(k+1)(k+2)\)\(= \dfrac{1}{4} k(k+1)(k+2)(k+3)\),
则当\(n=k+1\)时,
\(1×2×3+2×3×4+3×4×5+⋯+k(k+1)(k+2)+(k+1)(k+2)(k+3)\)
\(= \dfrac{1}{4} k(k+1)(k+2)(k+3)+(k+1)(k+2)(k+3)\)
\(=(k+1)(k+2)(k+3)( \dfrac{1}{4} k+1)\)
\(= \dfrac{1}{4} (k+1)[(k+1)+1][(k+1)+2][(k+1)+3]\),
所以当\(n=k+1\)时等式也成立.
综上所述,等式成立. -
证明 (1)当\(n=2\)时,左边 \(=1+\dfrac{1}{\sqrt{2}}=1+\dfrac{\sqrt{2}}{2}\),右边\(=\sqrt{2}\),等式成立.
(2)假设当\(n=k\)\((k≥2\)且\(k∈N^*)\)不等式成立,即\(1+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+\cdots+\dfrac{1}{\sqrt{k}}>\sqrt{k}\),
当\(n=k+1\)时, \(1+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+\cdots+\dfrac{1}{\sqrt{k}}+\dfrac{1}{\sqrt{k+1}}>\sqrt{k}+\dfrac{1}{\sqrt{k+1}}=\dfrac{\sqrt{k(k+1)}+1}{\sqrt{k+1}}\)\(>\dfrac{\sqrt{k \cdot k}+1}{\sqrt{k+1}}=\dfrac{k+1}{\sqrt{k+1}}=\sqrt{k+1}\),
\(\therefore\)当\(n=k+1\)时,不等式也成立.
\(\therefore\) 对\(n≥2\),\(n∈N^*\)时, \(1+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+\cdots+\dfrac{1}{\sqrt{n}}>\sqrt{n}\). -
答案 (1) \(a_1=1\),\(a_2=3\),\(a_3=7\),\(a_n=2^n-1\);(2) 略.
解析 (1)当\(n=1\)时,\(a_1=S_1=2a_1-1\),解得\(a_1=1\),
又\(S_{n+1}=2a_{n+1}-n-1\),\(S_n=2a_n-n\),两式相减可得
于是\(a_{n+1}=S_{n+1}-Sn=2a_{n+1}-n-1-2a_n+n\),
化为\(a_{n+1}=2a_n+1\),
从而可以得到\(a_2=3\),\(a_3=7\),
猜想通项公式\(a_n=2^n-1\);
(2)下面用数学归纳法证明:\(a_n=2^n-1\).
①当\(n=1\)时,\(a_1=1\)满足通项公式;
②假设当\(n=k\)时,命题成立,即a_k=2^k-1,
由(1)知 \(a_{k+1}=2 a_k+1=2\left(2^k-1\right)+1\), \(a_{k+1}=2^{k+1}-1\),
可得当\(n=k+1\)时命题成立.
由①②可证\(a_n=2^n-1\)成立. -
证明 ①当\(n=1\)时, \(2^{1+2} \cdot 3^1+5 \times 1-4=25\),能被\(25\)整除,命题成立.
②假设\(n=k\)\((k∈N^*)\)时, \(2^{k+2} \times 3^k+5 k-4\)能被\(25\)整除.
那么\(n=k+1\)时,原式 \(=2^{k+3} \times 3^{k+1}+5(k+1)-4\)
\(=6 \times 2^{k+2} \times 3^k+5(k+1)-4\)
\(=6\left[\left(2^{k+2} \times 3^k+5 k-4\right)-5 k+4\right]+5(k+1)-4\)
\(=6\left(2^{k+2} \times 3^k+5 k-4\right)-30 k+24+5 k+5-4\)
\(=6\left(2^{k+2} \times 3^k+5 k-4\right)-25(k-1)\).
\(\because 6\left(2^{k+2} \times 3^k+5 k-4\right)\)、\(-25(k-1)\)能被\(25\)整除,
\(\therefore n=k+1\)时,命题成立.
综上,对一切\(n∈N^*\),都有\(2^{n+2}×3^n+5n-4\)能被\(25\)整除. -
答案 (1) \(a_2= \dfrac{1}{3}\) , \(a_3=\dfrac{2}{7}\),\(a_4= \dfrac{1}{4}\);(2) \(a_n=\dfrac{2}{n+4}\),证明略 .
解析 (1)数列\(\{a_n\}\)满足 \(a_1=\dfrac{2}{5}\),\(a_{n+1} a_n+2a_{n+1}=2a_n\),\((n∈N^*)\).
\(n=1\)时,\(a_2= \dfrac{1}{3}\) ,\(n=2\)时,解得 \(a_3=\dfrac{2}{7}\),\(n=3\)时,解得\(a_4= \dfrac{1}{4}\).
(2)猜想:\(a_n=\dfrac{2}{n+4}\).
证明:①当\(n=1\)时, \(a_1=\dfrac{2}{5}=\dfrac{2}{1+4}\),猜想成立;
②假设当\(n=k\)\((k∈N^*)\)时猜想成立,即 \(a_k=\dfrac{2}{k+4}\).
那么,依题可得 \(a_{k+1}=\dfrac{2 a_k}{a_k+2}=\dfrac{2 \cdot \dfrac{2}{k+4}}{\dfrac{2}{k+4}+2}=\dfrac{2}{k+5}=\dfrac{2}{(k+1)+4}\).
所以,当\(n=k+1\)时猜想成立.
根据①和②,可知猜想对任何\(n∈N^*\)都成立. -
答案 (1) \(1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+\ldots+\dfrac{1}{(n+1)^2}<\dfrac{2(n+1)-1}{n+1}\);(2) 略.
解析 (1)第\(n\)个不等式为 \(1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+\ldots+\dfrac{1}{(n+1)^2}<\dfrac{2(n+1)-1}{n+1}\);
证明:(2)法一、数学归纳法:当\(n=1\)时,左边 \(=1+\dfrac{1}{2^2}=\dfrac{5}{4}<\dfrac{3}{2}=\)右边,不等式成立;
假设当\(n=k\)\((k∈N^*\)且\(k≥1)\)时不等式成立,
即\(1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+\ldots+\dfrac{1}{(k+1)^2}<\dfrac{2(k+1)-1}{k+1}\),
那么,当\(n=k+1\)时,
左边 \(=1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+\ldots+\dfrac{1}{(k+1)^2}+\dfrac{1}{(k+2)^2}<\dfrac{2(k+1)-1}{k+1}+\dfrac{1}{(k+2)^2}\)
\(<\dfrac{2 k+1}{k+1}+\dfrac{1}{(k+1)(k+2)}=\dfrac{2 k^2+k+4 k+2+1}{(k+1)(k+2)}=\dfrac{2 k^2+5 k+3}{(k+1)(k+2)}\)
\(=\dfrac{(k+1)(2 k+3)}{(k+1)(k+2)}=\dfrac{2[(k+1)+1]-1}{(k+1)+1}=\)右边,
\(\therefore\)当\(n=k+1\)时不等式成立.
综上所述,不等式 \(1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+\ldots+\dfrac{1}{(n+1)^2}<\dfrac{2(n+1)-1}{n+1}\)对于任意\(n∈N^*\)都成立;
法二、放缩法:\(\because n^2>n(n-1)\), \(\therefore \dfrac{1}{n^2}<\dfrac{1}{n(n-1)}\),
\(\therefore 1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+\ldots+\dfrac{1}{(n+1)^2}<1+\dfrac{1}{1 \times 2}+\dfrac{1}{2 \times 3}+\ldots+\dfrac{1}{n(n+1)}\)
\(=1+1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\ldots+\dfrac{1}{n}-\dfrac{1}{n+1}=2-\dfrac{1}{n+1}=\dfrac{2(n+1)-1}{n+1}\).
【B组---提高题】
1.用数学归纳法证明: \(|\sin n \alpha| \leq n|\sin \alpha|\) \((n∈N^*)\).
2.已知数列\(\{x_n\}\)满足\(x_1= \dfrac{1}{2}\), \(x_{n+1}=\dfrac{1}{1+x_n}\),\(n∈N^*\);
(1)猜想数列\(\{x_{2n}\}\)的单调性,并证明你的结论;
(2)证明:\(\left|x_{n+1}-x_n\right| \leq \dfrac{1}{6}\left(\dfrac{2}{5}\right)^{n-1}\).
3.已知数列\(\{a_n \}\)的各项都是正数,且满足\(a_0=1\),\(a_{n+1}= \dfrac{1}{2} a_n (4-a_n )\),\(n∈N\),
(1) 证明\(a_n<a_{n+1}<2\),\(n∈N\);\(\qquad \qquad\) (2) 求数列\(\{a_n \}\)的通项公式\(a_n\).
参考答案
-
证明 当\(n=1\)时,不等式的左边\(=|\sin α|\),右边\(=|\sin α|\),不等式成立;
假设\(n=k\)\((k∈N^*)\),\(|\sin kα|≤k|\sin α|\),
当\(n=k+1\)时,\(|\sin (k+1)α|=|\sin kα\cdot \cos α+\cos kα\cdot \sin α|\)
\(≤|\sin kα|\cdot |\cos α|+|\cos kα|\cdot |\sin α|≤|\sin kα|+|\sin α|≤(k+1)|\sin α|\),
即\(n=k+1\)时,不等式也成立.
综上可得,\(|\sin nα|≤n|\sin α|\) \((n∈N^*)\). -
答案 (1) 数列\(\{x_{2n}\}\)是递减数列 ;(2)略 .
解析 (1)由\(x_1= \dfrac{1}{2}\), \(x_{n+1}=\dfrac{1}{1+x_n}\),
\(\therefore x_2=\dfrac{2}{3}\), \(x_3=\dfrac{3}{5}\), \(x_4=\dfrac{5}{8}\), \(x_5=\dfrac{8}{13}\), \(x_6=\dfrac{13}{21}\),\(…\)
由\(x_2>x_4>x_6\)猜想:数列\(\{x_{2n}\}\)是递减数列
下面用数学归纳法证明:
(1)当\(n=1\)时,已证命题成立
(2)假设当\(n=k\)时命题成立,即 \(x_{2 k}>x_{2 k+2}\)
易知 \(x_{2 k}>0\),
那么\(x_{2 k+2}-x_{2 k+4}=\dfrac{1}{1+x_{2 k+1}}-\dfrac{1}{1+x_{2 k+3}}=\dfrac{x_{2 k+3}-x_{2 k+1}}{\left(1+x_{2 k+1}\right)\left(1+x_{2 k+3}\right)}\)
\(=\dfrac{x_{2 k}-x_{2 k+2}}{\left(1+x_{2 k}\right)\left(1+x_{2 k+1}\right)\left(1+x_{2 k+2}\right)\left(1+x_{2 k+3}\right)}>0\)
即 \(x_{2(k+1)}>x_{2(k+1)+2}\)
也就是说,当\(n=k+1\)时命题也成立,结合(1)和(2)知,命题成立
(2)当\(n=1\)时, \(\left|x_{n+1}-x_n\right|=\left|x_2-x_1\right|=\dfrac{1}{6}\),结论成立
当\(n≥2\)时,易知\(0<x_{n-1}<1\),
\(\therefore 1+x_{n-1}<2\), \(x_n=\dfrac{1}{1+x_{n-1}}>\dfrac{1}{2}\),
\(\therefore\left(1+x_n\right)\left(1+x_{n-1}\right)=\left(1+\dfrac{1}{1+x_{n-1}}\right)\left(1+x_{n-1}\right)=2+x_{n-1} \geq \dfrac{5}{2}\)
\(\therefore\left|x_{n+1}-x_n\right|=\left|\dfrac{1}{1+x_n}-\dfrac{1}{1+x_{n-1}}\right|=\dfrac{\left|x_n-x_{n-1}\right|}{\left(1+x_n\right)\left(1+x_{n-1}\right)} \leq \dfrac{2}{5}\left|x_n-x_{n-1}\right|\)
\(\leq\left(\dfrac{2}{5}\right)^2\left|x_{n-1}-x_{n-2}\right| \leq \cdots \leq\left(\dfrac{2}{5}\right)^{n-1}\left|x_2-x_1\right|\) \(=\dfrac{1}{6}\left(\dfrac{2}{5}\right)^{n-1}\). -
答案 (1)略 ;(2) \(a_n=2-\left(\dfrac{1}{2}\right)^{2^n-1}\) .
解析 (1)1°当\(n=1\)时,\(a_0=1\), \(a_1=\dfrac{1}{2} a_0\left(4-a_0\right)=\dfrac{3}{2}\) ,
\(\therefore a_0<a_1<2\),命题正确.
2°假设\(n=k\)时有 \(a_{k-1}<a_k<2\).
则\(n=k+1\)时, \(a_k-a_{k+1}=\dfrac{1}{2} a_{k-1}\left(4-a_{k-1}\right)-\dfrac{1}{2} a_k\left(4-a_k\right)\)
\(=2\left(a_{k-1}-a_k\right)-\dfrac{1}{2}\left(a_{k-1}^2-a_k^2\right)\)
\(=\dfrac{1}{2}\left(a_{k-1}-a_k\right)\left(4-a_{k-1}-a_k\right)\).
而 \(a_{k-1}-a_k<0\), \(4-a_{k-1}-a_k>0\), \(\therefore a_k-a_{k+1}<0\).
又 \(a_{k+1}=\dfrac{1}{2} a_k\left(4-a_k\right)=\dfrac{1}{2}\left[4-\left(a_k-2\right)^2\right]<2\)
\(\therefore n=k+1\)时命题正确.
由1°、2°知,对一切\(n∈N\)时有\(a_n<a_{n+1}<2\).
(2) \(a_{n+1}=\dfrac{1}{2} a_n\left(4-a_n\right)=\dfrac{1}{2}\left[-\left(a_n-2\right)^2+4\right]\),
所以 \(2\left(a_{n+1}-2\right)=-\left(a_n-2\right)^2\)
令\(b_n=a_n-2\),
则 \(b_n=-\frac{1}{2} b_{n-1}^2=-\frac{1}{2}\left(-\frac{1}{2} b_{n-2}^2\right)^2=-\frac{1}{2}\left(\frac{1}{2}\right)^2\left(b_{n-3}\right)^{2^2}=\cdots=-\left(\frac{1}{2}\right)^{1+2++2^{n-1}}\left(b_1\right)^{2^n}\),
又\(b_0=-1\),
所以 \(b_n=-\left(\dfrac{1}{2}\right)^{2^n-1}\),即 \(a_n=2+b_n=2-\left(\dfrac{1}{2}\right)^{2^n-1}\).
【C组---拓展题】
1.已知数列\(\{a_n \}\)中, \(a_1=\dfrac{1}{3}\), \(a_{n+1}=\sin \left(\dfrac{\pi}{2} a_n\right) \quad\left(n \in N^*\right)\),
(1)证明:\(0<a_n<a_{n+1}<1\);
(2)设数列\(\{a_n \}\)的前\(n\)项和为\(S_n\),证明 \(S_n>n-\dfrac{10}{3}\).
2.平面内\(n\)条直线,其中任何两条不平行,任何三条不共点.
(1)设这\(n\)条直线互相分割成\(f(n)\)条线段或射线,猜想\(f(n)\)的表达式并给出证明;
(2)求证:这\(n\)条直线把平面分成\(\dfrac{n(n+1)}{2}+1\)个区域.
参考答案
-
证明 (1)数学归纳法:
①当\(n=1\)时,\(a_1= \dfrac{1}{3}\) ,\(a_2= \dfrac{1}{2}\),显然有\(0<a_1<a_2<1\).
②假设当\(n=k\)\((k∈N^*)\),结论成立,即 \(0<a_k<a_{k+1}<1\),
那么 \(0<\dfrac{\pi}{2} a_k<\dfrac{\pi}{2} a_{k+1}<\dfrac{\pi}{2}\),
\(\therefore 0<\sin \left(\dfrac{\pi}{2} a_k\right)<\sin \left(\dfrac{\pi}{2} a_{k+1}\right)<1\),即 \(0<a_{k+1}<a_{k+2}<1\),
综上所述\(0<a_n<a_{n+1}<1\)成立.
(2)由(1)知: \(\dfrac{1}{3} \leq a_n<1\), \(\therefore 0<1-a_n \leq \dfrac{2}{3}\),
\(\therefore 0<\dfrac{\pi}{4}\left(1-a_n\right) \leq \dfrac{\pi}{6}\),
也即\(0<\sin \left[\dfrac{\pi}{4}\left(1-a_n\right)\right] \leq \sin \dfrac{\pi}{6}=\dfrac{1}{2}\) ;
\(1-a_n=1-\sin \left(\dfrac{\pi}{2} a_{n-1}\right)=1-\cos \left(\dfrac{\pi}{2}\left(1-a_{n-1}\right)\right)\),
\(=2 \sin ^2\left[\dfrac{\pi}{4}\left(1-a_{n-1}\right)\right] \leq \sin \left[\dfrac{\pi}{4}\left(1-a_{n-1}\right)\right]<\dfrac{\pi}{4}\left(1-a_{n-1}\right)\).
于是 \(1-a_n<\dfrac{\pi}{4}\left(1-a_{n-1}\right)<\cdots<\left(\dfrac{\pi}{4}\right)^{n-1}\left(1-a_1\right)=\left(\dfrac{\pi}{4}\right)^{n-1} \cdot \dfrac{2}{3}\),
得\(n-S_n<\dfrac{\dfrac{2}{3}\left(1-\left(\dfrac{\pi}{4}\right)^n\right)}{1-\dfrac{\pi}{4}}<\dfrac{\dfrac{2}{3}}{1-\dfrac{\pi}{4}}<\dfrac{\dfrac{2}{3}}{1-\dfrac{3.2}{4}}=\dfrac{\dfrac{2}{3}}{\dfrac{1}{5}}=\dfrac{10}{3}\),
故\(S_n>n-\dfrac{10}{3}\). -
答案 (1) \(f(n)=n^2\);(2) 略.
解析 (1)解:\(f(2)=4\),\(f(3)=9\),\(f(4)=16\),
\(\therefore\)猜想\(f(n)=n^2\).以下用数学归纳法证明:
①当\(n=2\)时,\(f(2)=4=2^2\),猜想正确.
②假设\(n=k\)\((k≥2)\)时猜想正确,即\(f(k)=k^2\),
则当\(n=k+1\)时,这第\(k+1\)条直线与原来的\(k\)条直线分别相交,新增\(k\)个交点,
它们分别把原来的一条线段或射线一分为二,
使原来的\(k\)条直线新分割出\(k\)条线段或射线,
又这\(k\)个交点还把第\(k+1\)条直线分割为\(k+1\)条线段或射线,
\(\therefore f(k+1)=f(k)+k+(k+1)=k^2+2 k+1=(k+1)^2\).
\(\therefore\) 当\(n=k+1\)时,猜想也正确.
根据①②知,对大于\(1\)的任意自然数\(n\),猜想都正确.
(2)证明:①当\(n=1\)时,一条直线把平面分为两部分,
而\(n=1\)时 \(\dfrac{n(n+1)}{2}+1=2\), \(\therefore n=1\)时命题正确.
②假设\(n=k\)时命题正确,
即\(k\)条直线把平面分成 \(k(k)=\dfrac{k(k+1)}{2}+1\)个区域,
则\(n=k+1\)时,第\(k+1\)条直线 \(l_{k+1}\)与原来的\(k\)条直线可交于\(A_1\),\(A_2\),\(…\),\(A_k\)共\(k\)个交点,截成\(k+1\)条线段或射线,而每一条线段或射线都把它们所占的一块区域一分为二,故新增加出\(k+1\)块区域,
因此\(k+1\)条直线把平面共分成 \(\dfrac{k(k+1)}{2}+(k+1)+1\),
即 \(\dfrac{(k+1)(k+2)}{2}+1\)个区域.
\(\therefore\) 当\(n=k+1\)时命题也成立.
由①②可知,对任意的\(n∈N^*\),命题都成立.