例子一、关于$x$的不等式$x^2-ax+a+3\ge 0$在区间$[-2,0]$上恒成立，则实数a的取值范围是$(\qquad\qquad )$
${\color{Red}法一：分参法\quad }$ (全分离):
$x^2+3\ge a(x-1)\quad \because x-1\lt 0\Rightarrow a\ge \cfrac{x^2+3}{x-1}$
设$f(x)=\cfrac{x^2+3}{x-1} {\color{Red}只需a\ge f(x)_{max}} $即可。
换元，令$t=x-1 \quad(-3\le t \le -1)\Rightarrow x=t+1$
$f(t)=\cfrac{(t+1)^2+3}{t} =\cfrac{t^2+2t+4}{t}=t+\cfrac{4}{t}+2 $
$=t+\cfrac{4}{t}+2 =-(-t-\cfrac{4}{t})+2\le -2\sqrt[]{(-t)\times(-\cfrac{4}{t} )}+2=-2 $
$f(t)\le -2\quad a\ge f(x)_{max}=f(t)_{max}=-2$
${\color{Red}法二 ：函数法}$
$令f(x)=x^2-ax+a+3$
${\color{Green}\quad ①\quad } 当\cfrac{a}{2}\le -2,即a\le -4时，f(x)_{min}=f(-2)=7+3a\ge \Rightarrow a\ge -\cfrac{7}{3} $ 矛盾,无解。
${\color{Green}\quad ②\quad} 当\cfrac{a}{2}\ge 0\Rightarrow a\ge 0,f(x)_{min}=f(0)=a+3\ge 0\Rightarrow a\ge -3 \quad\therefore {\color{Purple}a\ge 0 } $
${\color{Green} \quad③\quad} -2\lt \cfrac{a}{2} \lt 0,即-4\lt a\lt 0，f(x)_{min}=f(\cfrac{a}{2})$
$=-\cfrac{a^2}{4} +a+3\ge 0\Rightarrow (a+2)(a-5)\le 0\Rightarrow {\color{Purple} -2\le a\le 6} $
$\therefore \quad -2\le a\le 0$
综上①②③所述，${\color{Purple} a\ge -2 }$
${\color{Red} 法三：图像法\quad}$半分离法
$x^2-ax+a+3\ge 0\Leftrightarrow x^2+3\ge a(x-1)$
$y_1=x^2+3\quad y_2= a(x-1)\quad$ 抛物线在直线上方，相切时取等号
直线$y_2= a(x-1)恒过(1,0)，a$为直线斜率，切线斜率为正，直线小于等于切线斜率，直线恒在抛物线下方；直线斜率为负，大于等于切线斜率，直线均恒在抛物线下方。（此题只讨论在抛物线左侧[-2,0]范围）
联立抛物线与直线方程，$x^2-ax+a+3=0\Rightarrow \bigtriangleup =a^2-4(a+3)，解得a=-2或a=6(右侧舍去)$
所以$a\ge -2$

题目2：关于x的不等式$kx-ln x\ge 0$恒成立，求实数k的取值范围$(\qquad)$
${\color{Red}法一：分参法\quad }$ (全分离):
$ kx-\ln x\ge 0\Rightarrow kx\ge\ln x\Rightarrow {\color{Red} k\ge (\cfrac{\ln x}{x} )_{max}}$
$k\ge \cfrac{\ln x}{x},令f(x)= \cfrac{\ln x}{x},{f}'(x)=\cfrac{1-\ln x}{x^2},$
${\color{Green} \quad ①\quad} (0,e),{f}' (x)\gt 0,f(x)\nearrow $
${\color{Green} \quad ②\quad (e,+\infty )} ,{f}' (x)\lt 0,f(x)\searrow$
故$f(x)在x=e处有极大值，f(x)\le f(x)_{max}=f(e)=\cfrac{1}{e}\Rightarrow k\ge \cfrac{1}{e}$
${\color{Red}法二 ：函数法}$
$kx-\ln x\ge 0;令f(x)=kx-\ln x;{f}' (x)=k-\cfrac{1}{x}$
${\color{Green}\quad ①} 当k\le 0,{f}' (x)\lt 0,f(x)\searrow $
$f(x)$无极大值。故无解。
${\color{Green}\quad ② } k\gt0时，{f}' (x)=k-\cfrac{1}{x}=\cfrac{kx-1}{x} $;
${\color{Purple} \qquad 1^\circ } 当0\lt x\lt \cfrac{1}{k}时， {f}' (x)\lt 0,f(x)\searrow $
${\color{Purple} \qquad 2^\circ } 当 x\gt \cfrac{1}{k}时， {f}' (x)\gt 0,f(x)\nearrow $
故$f(x)在x=\cfrac{1}{k}有极小值f(x)_{min}=f(\cfrac{1}{k})=1+\ln k $
$1+\ln k \ge 0\Rightarrow k\ge \cfrac{1}{e} $
${\color{Red} 法三：图像法\quad}$半分离法
$kx\ge \ln x$
即左边的过原点的直线恒在对数函数左上方。当且仅当相切时取=,因而变成求过原点与对数函数相切的直线问题。
$f(x)=\ln x\Rightarrow {f}'(x)=\cfrac{1}{x},设切点为(m,\ln m), {f}' (m)=\cfrac{1}{m} $
$切线方程为：y-\ln m=\cfrac{1}{m}(x-m) $
将$(0,0)$代入切线方程：$\ln m=1 \Rightarrow m=e ,故k\ge \cfrac{1}{m} =\cfrac{1}{e} $

总结：图像法仅适用于解答填空与选择题，分离适用于大题目，且简涪，但仅适用于能完全分离参数的题目；函数法适用于不能全部分离参数的题目，较繁

${\color{Green}1. f(x)=\cfrac{\ln x}{x} } 最常用，没有之一，{\color{Red} e定要记住！} \quad$
$低价比高价有{\color{Red}极大值 } ，导函数的符号函数是1-\ln x,极值为{\color{Red}(e,\frac{1}{e}) } ;$
$它的倒数函数{\color{Green}\quad 2、\quad g(x)=\cfrac{x}{\ln x} } 高价比低价有{\color{Red} } {\color{Red} 极小值。倒不变！(e,e)}) $
$它的积函数{\color{Green} \quad 3、\quad h(x)=x\ln x} ,{\color{Red} 有极小值，为倒反(\cfrac{1}{e} ,-\cfrac{1}{e} )}$

已知 $f(x)=e^x\ln x+\cfrac{2e^{x-1} }{x} ,求证:f(x)\gt 1$
已知 $f(x)=ae^x-\ln x-1,证:当a\ge \cfrac{1}{e}时 ,f(x)\ge 0$

${\color{Red} 分式裂项}$ 分式裂项
一、基本原理
裂消原理：$\cfrac{大－小}{小*大}＝\cfrac{1}{小}－\cfrac{1}{大}$
相消规律：对称性
相邻项，首尾各剩下1项；
隔一项，首尾各剩下2项；
隔两项，首尾各剩下3项；
二、八大裂项模型：
${\color{Red} \quad ①\quad a_n=\cfrac{1}{n(n+1)} =\cfrac{1}{n} -\cfrac{1}{n+1} } \quad \qquad$
${\color{Green}\quad ②\quad a_n=\cfrac{1}{(2n-1)(2n+1)} =\cfrac{1}{2} \cfrac{(2n+1)-(2n-1)}{(2n-1)(2n+1)}=\cfrac{1}{2}\cdot (\cfrac{1}{2n-1}-\cfrac{1}{2n+1})}\quad \qquad$
${\color{Red} \quad ③\quad a_n=\cfrac{3}{(6n-1)(6n+5)} =\cfrac{1}{2} \cfrac{(6n+5)-(6n-1)}{(6n-1)(6n+5)}=\cfrac{1}{2} \cfrac{6}{(6n-1)(6n+5)}}\quad \qquad$
${\color{Green}\quad ④\quad a_n=\cfrac{2^{n+1}}{(2^n+1)(2^{n+1}+1)} =\cfrac{2[(2^{n+1}+1)-(2^n+1)]}{(2^n+1)(2^{n+1}+1)}}\quad $
${\color{Red} \quad ⑤\quad a_n=\cfrac{n+1}{4n^2(n+2)^2} =\cfrac{1}{4} \cdot \cfrac{1}{4} \cdot \cfrac{(n+2)^2-n^2}{n^2(n+2)^2}}\quad $

${\color{Green}\quad ⑥\quad a_n=(-1)^n\cfrac{2n+1}{n(n+1)}=(-1)^n\cdot\cfrac{n+1+n}{n(n+1)} }\quad 裂项要出现负号，这里有-1$
${\color{Red} \quad ⑦\quad a_n=\cfrac{1}{n(n+1)(n+2)} =\cfrac{1}{2} \cfrac{(n-2)-n}{n(n+1)(n+2)}=\cfrac{1}{2} [\cfrac{1}{n(n+1)}-\cfrac{1}{(n+1)(n+2)} ]}\quad$当中间n+1项是透明的
${\color{Green}\quad ⑧\quad a_n=\cfrac{n}{(n+1)!}=\cfrac{(n+1)-1}{1\times 2\times 3\times4 \times \cdots \cdots (n+1)} ＝\cfrac{1}{n!}-\cfrac{1}{(n+1)!} }\quad$ 只看首尾项
实战：

(1)设$b_n=(2n-1)a_n\quad S_n表示b_n的前n项和，S_n=2n$
$①n=1时，S_1=b_1=2;$
$②n\ge 2时，S_n-S_{n-1}=b_n=2n-2(n-1)=2,故b_n$是常数数列。
$b_n=(2n-1)a_n=2\Rightarrow a_n=\cfrac{2}{2n-1}$
(2)$\cfrac{a_n}{2n+1} =\cfrac{2}{(2n-1)(2n+1)}=\cfrac{1}{2n-1}-\cfrac{1}{2n+1}$
设前n项和为$T_n={\color{Green} \cfrac{1}{1}} -\cfrac{1}{3}+\cfrac{1}{3} -\cfrac{1}{5}+\cfrac{1}{5} -\cfrac{1}{7}+\cdots \cdots+$
$\cfrac{1}{2n-3}-\cfrac{1}{2n-1}+\cfrac{1}{ 2n-1}-{\color{Green} \cfrac{1}{2n+1}} =\cfrac{2n}{2n+1}$

$n\ge 2时，a_n=S_n-S_{n-1}=\sqrt{S_n} +\sqrt{S_{n-1}} \Rightarrow (\sqrt{S_n})^2 -(\sqrt{S_{n-1}})^2 =\sqrt{S_n} +\sqrt{S_{n-1}}$
$\sqrt{S_n} -\sqrt{S_{n-1}}=1\Rightarrow \sqrt{S_n}$ 是公差为1的首项为$\sqrt{a_1} $的等差数列。
$\sqrt{S_n} =n\Rightarrow S_n=n^2\quad a_n=n^2-(n-1)^2=2n-1$
$b_n=(-1)^n\cfrac{n}{a_na_{n+1}}= (-1)^n\cfrac{n}{(2n-1)(2n+1)}=\cfrac{1}{4} \times (-1)^n\cfrac{2n+1+2n-1}{(2n-1)(2n+1)}$
$b_n=\cfrac{1}{4} (-1)^n(\cfrac{1}{2n-1}+\cfrac{1}{2n+1} )$
$T_{2n}=\cfrac{1}{4} [{\color{Green}-(\cfrac{1}{1} } +\cfrac{1}{3})+(\cfrac{1}{3} +\cfrac{1}{5})-(\cfrac{1}{5}-\cfrac{1}{7})+\dots +\cfrac{1}{4n-1}+{\color{Green}\cfrac{1}{4n+1} } ]$
$T_{2n}=-\cfrac{n}{4n+1}$

${\color{Red} PART\quad TWO\quad 根式裂项}$
一、模型总结：
${\color{Red} \quad ②\quad a_n=\cfrac{1}{\sqrt{n}+\sqrt{n+1} } =\cfrac{\sqrt{n+1}-\sqrt{n} }{(\sqrt{n+1}+\sqrt{n} )(\sqrt{n+1}-\sqrt{n} )} =\sqrt{n+1}-\sqrt{n} }$
${\color{Green}\quad ②\quad a_n=\cfrac{1}{\sqrt{n}(n+1)+n\sqrt{n+1} } =\cfrac{1}{\sqrt{n}\sqrt{n+1}(\sqrt{n+1} +\sqrt{n} ) } =\cfrac{1}{\sqrt{n} \sqrt{n+1} }(\sqrt{n+1}-\sqrt{n})}$
${\color{Green} \quad ②\quad a_n=\cfrac{1}{\sqrt{n} } -\cfrac{1}{\sqrt{n+1}} } $


${\color{Green} a_n=\cfrac{1}{\sqrt{n} } -\cfrac{1}{\sqrt{n+1}} } $
$S_n=\cfrac{1}{\sqrt{1} } -\cfrac{1}{\sqrt{n+1}} $
$S_{2024}=1-\cfrac{1}{\sqrt{2024+1}} ,45^2=2025,2^2=3+12^2+....+45^2,45-2+1=44个$

${\color{Red} 第三部分、平方递推裂项} $
一、模型总结：
${\color{Green} ①a_{n+1}=a_n^2-a_n+1,(a_{n+1}=a_n^2两边对数求通项，但此式求不了通项。} $
$a_{n+1}-a_n=(a_n^2-1)^2,a_1\ne 1时，$是递增数列。
$a_{n+1}=a_n^2-a_n+1\Rightarrow a_{n+1}=a_n(a_n-1)+1\Rightarrow a_{n+1}-1=a_n(a_n-1)\Rightarrow 取倒\cfrac{1}{a_{n+1}-1}=\cfrac{1}{a_n(a_n-1)} $
$\cfrac{1}{a_{n+1}-1}=\cfrac{1}{a_n(a_n-1)} =\cfrac{1}{a_n-1} -\cfrac{1}{a_n} \Rightarrow \cfrac{1}{a_n}=\cfrac{1}{a_n-1} - \cfrac{1}{a_{n+1}-1}$裂项完成
步骤：因式分解；移系数；取倒数；裂项；互移。

$\cfrac{1}{a_n}=\cfrac{1}{a_n-1} - \cfrac{1}{a_{n+1}-1}$
取倒前的那一步：$a_{n+1}-1=a_n(a_n-1)\Rightarrow \cfrac{1}{a_n}=\cfrac{a_n-1}{a_{n+1}-1} $
$A_n=1-\cfrac{1}{a_{n+1}}\quad B_n=\cfrac{a_1-1}{a_{n+1}-1} $

步骤：因式分解；移系数；取倒数；裂项；互移。
$ a_{n+1}-2=\cfrac{1}{2} a_n(a_n-2)\Rightarrow \cfrac{1}{a_{n+1}-2} =\cfrac{2}{ a_n(a_n-2)} $
$ \cfrac{1}{a_{n+1}-2} =\cfrac{a_n-(a_n-2)}{a_n(a_n-2)} =\cfrac{1}{a_n-2}-\cfrac{1}{a_n} $
$\Rightarrow \cfrac{1}{a_n} =\cfrac{1}{a_n-2}-\cfrac{1}{a_{n+1}-2} $
$a_1=\cfrac{5}{2} \quad \cfrac{1}{a_1}+ \cfrac{1}{a_2}+...+\cfrac{1}{a_{2020}} \quad $
$=(\cfrac{1}{a_1 -2}-\cfrac{1}{a_2-2})+(\cfrac{1}{a_2-2} -\cfrac{1}{a_3-2})+\cdots +(\cfrac{1}{a_{2020}-2}-\cfrac{1}{a_{2021}-2} )$
$\Rightarrow =\cfrac{1}{\cfrac{5}{2} -2}-\cfrac{1}{a_{2021}-2} =2-\cfrac{1}{a_{2021}-2}$
$a_{n+1}-a_n=\cfrac{1}{2}(a_n-2)^2\gt0 , a_n\nearrow 易证a_{2023}\gt 3$

${\color{Red} PART \quad4　\quad三角裂项 }$
${\color{Red}一、模型总结：来自差角公式：}$
${\color{Red} \quad①\quad a_n=\tan n\cdot \tan (n+1)} $
$\tan 1=\tan [(n+1)-n]=\cfrac{\tan (n+1)-\tan n}{1+\tan n\cdot \tan (n+1)} \Rightarrow $
$1+\tan n\cdot \tan (n+1)=\cfrac{\tan (n+1)-\tan n}{\tan 1} \Rightarrow 1+\tan n\cdot \tan (n+1) =\cfrac{1}{\tan 1} [\tan (n+1)-\tan n]$
${\color{Red} \tan n\cdot \tan (n+1) =\cfrac{1}{\tan 1} [\tan (n+1)-\tan n]-1} $
${\color{Red} S_n =\cfrac{1}{\tan 1} [\tan (n+1)-\tan 1]-n} $
${\color{Green} \quad ②\quad a_n=\cfrac{1}{\sin n\sin (n+1)} } $
${\color{Green} \because \sin 1}=\sin[(n+1)-n]=\sin (n+1)\cos n-\cos (n+1)\sin n$
右边分子分母$\times \sin 1 \quad \therefore $
$a_n=\cfrac{\sin 1}{\sin 1\sin n\sin (n+1)}=\cfrac{1}{\sin 1}\cdot \cfrac{\sin (n+1)\cos n-\cos (n+1)\sin n}{\sin n\sin (n+1)} $
${\color{Green} a_n= \cfrac{1}{\sin n\sin (n+1)} } =\cfrac{1}{\sin 1}\cdot [\cfrac{1}{\tan n} -\cfrac{1}{\tan (n+1)} ]$
${\color{Red}\quad ③\quad a_n=\cfrac{1}{\cos n\cos (n+1)} } $
同上，分子分母$\times \sin1 \quad a_n=\cfrac{1}{\sin1} \cdot\cfrac{\sin1}{\cos n\cos (n+1)} =$
$a_n=\cfrac{1}{\sin1} \cdot\cfrac{\sin (n+1)\cos n-\cos (n+1)\sin n}{\cos n\cos (n+1)} =\cfrac{1}{\sin1} \cdot[\tan (n+1)-\tan n]$
${\color{Red}\quad a_n=\cfrac{1}{\sin1} \cdot[\tan (n+1)-\tan n]} $

$7a_4=63\Rightarrow a_4=9,a_5=9,d=a_5-a_4=3,a_1=0,a_n=3n-3$
$b_n=\cfrac{\sin 3}{\cos(3n-3)\cos (3n)} =\cfrac{\sin [3n-(3n-3)]}{\cos(3n-3)\cos (3n)} =\cfrac{\sin 3n\cos(3n-3)-\cos 3n \sin (3n-3)}{\cos(3n-3)\cos (3n)}=\tan 3n -\tan (3n-3)$
$Ｓ_n=b_1+b_2+\dots +b_n =\tan 3 -{\color{Red} \tan 0} +\tan 6-\tan 3+\dots +{\color{Red} \tan 3n} -\tan (3n-3)\quad$位置对称！！
$S_n=\tan 3n$

抽象数列：
${\color{Green} 1、a_{n+m}=a_n+a_m{\color{Orange} \Rightarrow a_n=na_1=nd} }; $
${\color{Red} 2、a_{n+m}=a_na_m\Rightarrow a_n=a_1^n=q^n} $
${\color{Blue} 3、a_{m+n}+a_{m-n}=2a_m+2a_n\Rightarrow a_n=n^2} $
${\color{Tan} 4、a_{m+n}+a_{m-n}=2a_ma_n\Rightarrow a_n=\cos (\omega n)} $

$q^9=27\therefore q^3=3, \cfrac{a_1a_5+a_3a_6}{a^2_6+2a_3^2}$
$=\cfrac{(q^3)^2+q^3q^6}{(q^6)^2+2(q^3)^3} =\cfrac{q^6(1+q^3)}{q^6(q^6+2q^3)} =\cfrac{4}{15}$

令$m=1,d=a_1=2,S_{k+2}-S_k=a_{k+1}+a_{k+2}=26,k=6$

令$q=1,a_1=q=2$

$\{a_n\}=n^2,求S_n\quad$
$n^3-(n-1)^3=n^3-(n^3-3n^2+3n-1)=3n^2-3n+1\quad $
$(n-1)^3-(n-2)^3=3(n-1)^2-3(n-1)+1\quad $
$(n-2)^3-(n-3)^3=3(n-2)^2-3(n-2)+1\quad $
$(n-3)^3-(n-4)^3=3(n-3)^2-3(n-3)+1\quad $
$\dots \dots \dots $
$2^3-1^3=3\cdot2^2-3\cdot 2+1\quad $
$1^3-0^3=3\cdot1^2-3\cdot 1+1\quad $
左右两边相加，$n^3=3S_n-3\cfrac{n(n+1)}{2}+n$
$2n^3=6S_n-3n(n+1)+2n=6S_n-3n^2-n\Rightarrow 6S_n=2n^3+3n^2+n$
$6S_n=n(2n^2+3n+1)=n(n+1)(2n+1)\Rightarrow S_n=n(2n^2+3n+1)=\cfrac{1}{6}\cdot n(n+1)(2n+1)$
$S_7=7\times 8\times 15{\div} 6=140$

令$a_n=\cos (\omega n)\Rightarrow a_1=\cos \omega =\cfrac{1}{2} $
$\therefore \omega =\cfrac{\pi}{3} ,T=\cfrac{2\pi }{\omega }=6$
$2023=337\times 6+1,a_1=a_{2023}=\cfrac{1}{2} $