1、证明：$e^x\gt \ln (x+3)-\cfrac{1}{2}$
$\Leftrightarrow e^x-\ln (x+3)+\cfrac{1}{2}\gt 0\Leftrightarrow {\color{Red} [e^x-\ln (x+3)+\cfrac{1}{2}]_{min}\gt 0} $
设$f(x)=e^x-\ln (x+3)+\cfrac{1}{2}\quad {f}'(x)=e^x-\cfrac{1}{x+3} \nearrow $
${\color{Red} 观察} \quad {f}' (0)=e^0 -\cfrac{1}{3} =\cfrac{2}{3}\gt0,\quad {f}' (-1)=\cfrac{1}{e}-\cfrac{1}{2} \lt 0，记住： \cfrac{1}{e}=0.36788$
$\exists x_0\in (-1,0)使得{f}' (x_0)=e^{x_0}-\cfrac{1}{x_0+3} =0\Rightarrow e^{x_0}=\cfrac{1}{x_0+3,} $
$(-3,x_0),{f}' (x)\lt 0,f(x)\searrow ;$
$(x_0,+\infty),{f}' (x)\gt 0,f(x)\nearrow ;{\color{Red} 即f(x)在x_0处有最小值f(x_0)} $
$f(x)\ge f(x_0)=e^{x_0}-\ln(x_0+3)+\cfrac{1}{2}=\cfrac{1}{x_0+3}+x_0+\cfrac{1}{2}$
判断$f(x_0)\ge 0$有两种方法：利用导数和对勾函数；
方法1：
$\because x_0\in (-1,0),且f(x_0)=\cfrac{1}{x_0+3}+x_0+3-3+\cfrac{1}{2}在(-2,+\infty)$单调递增，$f(x_0)\gt f(-1)=0$
$\therefore f(x)\ge f(x_0)\gt 0$
方法2：
令$x\in (-1,0)，设g(x)=\cfrac{1}{x+3}+x+\cfrac{1}{2} \quad {g}'(x)= 1-\cfrac{1}{(x+3)^2}\gt 0$
$\therefore \quad \Rightarrow g(x)\nearrow g(x)\gt g(-1)=\cfrac{1}{-1+3}-1+\cfrac{1}{2} =0$

2、证明:$x^2e^x\gt \ln x+\cfrac{1}{3}$
令$f(x)=x^2e^x- \ln x-\cfrac{1}{3} \quad (x\gt0 ) {\color{Red}\quad 即证f(x)_{min}\gt 0}$
${f}' (x)=(x^2+2x)e^x-\cfrac{1}{x}\quad$
设$g(x)= {f}' (x)\quad {g}' (x)=(x^2+2x+2x+2)e^x+\cfrac{1}{x^2}\gt 0 $
$\because {g}' (x)\gt 0\Rightarrow {f}' (x)\nearrow$
${f}' (1)\gt 0\quad {f}' (\cfrac{1}{5} )\lt 0$
$\exists x_0\in (\cfrac{1}{5},1),使得{f}' (x_0)=(x_0^2+2x_0)e^{x_0}-\cfrac{1}{x_0} =0$
即$x\in (0,x_0)\quad {f}' (x)\lt 0\quad f(x)\searrow $
$x\in (x_0.+\infty)\quad {f}' (x)\gt 0\quad f(x)\nearrow $
$f(x)\ge f(x_0)=x_0^2e^{x_0}- \ln x_0-\cfrac{1}{3}= {\color{Red}f(x)_{min}}$
$x_0(x_0+2)e^{x_0}=\cfrac{1}{x_0} \Rightarrow x_0^2e^{x_0}=\cfrac{1}{x_0+2}$
${\color{Red}f(x)_{min}}=x_0^2e^{x_0}- \ln x_0-\cfrac{1}{3}=\cfrac{1}{x_0+2}-\ln x_0-\cfrac{1}{3}$
设$g(t)=\cfrac{1}{t+2}-\ln t-\cfrac{1}{3}\quad (\cfrac{1}{5}\lt t \lt 1)$
$g(t)单调递减，设g(t)\gt g(1)=\cfrac{1}{1+2}-\ln 1-\cfrac{1}{3} =0＝ {\color{Red}f(x)_{min}}\quad $得证

３、证明：当$a\ge 1时，3e^x+ax^2+x-1\gt 0$
${\color{Red} 方法一:隐零点回避} a\ge 1\quad \Rightarrow ax^2\ge x^2,\quad 3e^x+ax^2+x-1\gt 0\Leftrightarrow 3e^x+ax^2+x-1\ge 3e^x+x^2+x-1\gt 0$
$3e^x+x^2+x-1\gt 0\Leftrightarrow 3+\cfrac{x^2+x-1}{e^x} \gt0\quad \Leftrightarrow {\color{Red} 证明: (3+\cfrac{x^2+x-1}{e^x})_{min}\gt0}$
令$f(x)=3+\cfrac{x^2+x-1}{e^x}$
${f}' (x)=\cfrac{2x+1-(x^2+x-1)}{e^x} =\cfrac{-(x^2-x-2)}{e^x}=\cfrac{-(x-2)(x+1)}{e^x} $
$x\in (-\infty ,-1)\quad {f}' (x)\lt 0\quad f(x)\searrow $
$x\in (-1,2)\quad {f}' (x)\gt 0\quad f(x)\nearrow ;$
$x\in (2,+\infty)\quad {f}' (x)\lt 0\quad f(x)\searrow $
当$x\gt 2时，f(x)单调递减，但x^2+x-1\gt 0即，f(x)\gt 3$,所以$f(x)的最小值为f(-1)=3-e\gt 0$恒成立。
${\color{Red} 方法二:} 设f(x)=3e^x+x^2+x-1\quad {\color{Red} 证明f(x)_{min}\gt 0}\quad {f}' (x)=3e^x+2x+1$
${f}' (x)=3e^x+2x+1\nearrow {f}'(-1)=\cfrac{3}{e}-2+1\gt 0 \qquad {f}' (-2)=4\lt 0$
故$\exists x_0\in (-2,-1),使得{f}'(x_0)=0 即，3e^{x_0}=-2x_0-1$
$f(x)在(-\infty,x_0)\searrow ;f(x)在(x_0,+\infty)\nearrow$
故$f(x)\ge f(x_0)=3e^{x_0}+x_0^2+x_0-1=-2x_0-1+x_0^2+x_0-1=x_0^2-x_0-2=(x_0-2)(x_0+1)$
$\because x_0\lt -1,\therefore (x_0-2)\lt 0且 x_0+1\lt 0,故(x_0-2)(x_0+1)\gt 0\Rightarrow f(x)\gt 0得证$

4、证明:当$a\ge 1时，a(x+1)\gt \cfrac{\ln x+1}{x}$
${\color{Red} 法一: } a\ge 1\Rightarrow a(x+1)\ge x+1\gt \cfrac{\ln x+1}{x}$
如果你很熟悉那6个超越函数的极值的话，一眼便看出右边有极大值。若看不出来，就证左边的最小值大于右边的最大值。
$\cfrac{e\ln (ex)}{ex}= e\cdot \cfrac{1}{e} =1，而左边的函数单调递增，当x\gt 0,左边\gt 1,$得证
${\color{Red} 方法二：} a(x+1)\ge x+1\gt \cfrac{\ln x+1}{x}\Rightarrow x+1-\cfrac{\ln x+1}{x}\gt 0
\quad设f(x)=x+1-\cfrac{\ln x+1}{x}$
${\color{Red} 即证f(x)_{min} \gt 0}\quad {f}' (x)=1-\cfrac{-\ln x}{x^2} =\cfrac{x^2+\ln x}{x^2} $
而$g(x)=x^2+\ln x在(0,+\infty)\nearrow 且g(1)=1\gt 0\quad g(\cfrac{1}{e})=\cfrac{1}{e^2}-1\lt 0 $
$\exists x_0\in (\cfrac{1}{e},1),使得g(x_0)=x_0^2+\ln x_0=0$
$f(x)在(0,x_0)\searrow ,(x_0,+\infty)\nearrow ，f(x)\ge {\color{Red} f(x)_{min} } =f(x_0)=x_0+1-\cfrac{\ln x_0+1}{x_0} =$
$x_0+1-\cfrac{- x_0^2+1}{x_0} =2x_0-\cfrac{1}{x_0} +1=g(x_0)\nearrow $
$g(x_0)的最大值为g(1)=0，\therefore f(x)\ge g(1)\gt f(x_0)= {\color{Red} f(x)_{min} } $
${\color{Red} 法三：对数单身狗：隐零点回避}$
$x+1\gt\cfrac{\ln x+1}{x}\Leftrightarrow {\color{Red} x^2+x\gt \ln x+1} \Rightarrow x^2+x-\ln x-1\gt 0$
令$f(x)= x^2+x-\ln x-1\quad {f}' (x)=2x+1-\cfrac{1}{x} =\cfrac{2x^2+x-1}{x}=\cfrac{(2x-1)(x+1)}{x} $
$x\in (0,\cfrac{1}{2}),{f}' (x)\lt 0,f(x)\searrow ;\qquad$
$x\in (\cfrac{1}{2},+\infty),{f}' (x)\gt 0,f(x)\nearrow \Rightarrow f(x)在x=\cfrac{1}{2}$ 有极小值。
$f(x)\ge f(1)=\ln 2-\cfrac{1}{4}\gt 0，$得证。

5、已知函数$f(x)=ax^2-ax-x\ln x,且f(x)\ge 0 $(2017年全国2－17题)
(1)求$a$;
(2)证明：$f(x)$存在唯一的极大值点$x_0$，且有$e^{-2}\lt f(x_0)\lt 2^{-2}$
解：(1)$ax^2-ax-x\ln x\ge 0\Leftrightarrow a(x-1)\ge \ln x$
左边为过$(1,0)$的直线$y=a(x-1)$；右边为对数函数，当且仅当直线是对数的切线时，$f(x)\ge 0$成立，易解得$a=1$;
(2)证明：$f(x)=x^2-x-x\ln x存在极大值点x_0$
$x^2-x-x\ln x=0$
${f}' (x)=2x-1-(\ln xx+1)=2x-2-\ln x\quad (x\gt0)$
令$g(x)={f}' (x)\quad {g}' (x)=2-\cfrac{1}{x} =\cfrac{2x-1}{x} \quad (x\gt 0 )$
故${f}' (x)在(0,\cfrac{1}{2})\searrow (\cfrac{1}{2} ,+\infty ) \nearrow $
${f}' (\cfrac{1}{e})=\cfrac{2}{e}-2-\ln \cfrac{1}{e}=\cfrac{2}{e}-1\lt 0, $
${f}' (\cfrac{1}{e^2})=\cfrac{2}{e^2}-2-\ln \cfrac{1}{e^2}=\cfrac{2}{e^2}\gt 0$
故$\exists x_0\in (\cfrac{1}{e^2},\cfrac{1}{e} )，使得{f}' (x_0)={\color{Red} 2x_0-2-\ln x_0=0} $
又$\because {f}' (1)=0,故{f}' (x)在 (0,x_0)大于0，(x_0,1)小于0，(1,+\infty)大于0$
$\therefore f(x)$存在唯一极大值点$x_0,\Rightarrow f(x_0)\gt f(\cfrac{1}{e})=\cfrac{1}{e^2}-\cfrac{1}{e}-\cfrac{1}{e}\ln \cfrac{1}{e}=\cfrac{1}{e^2}$
${\color{Green} 放大}f(x_0)=x_0^2-x_0-x_0\ln x_0= x_0^2-x_0-x_0{\color{Red} (2x_0-2) }=-x_0^2+x_0\lt -(\cfrac{1}{2})^2+\cfrac{1}{2}=\cfrac{1}{4}$
$\therefore e^{-2}\lt f(x_0)\lt 2^{-2}$

$6、2015年全国1卷21题目；设函数f(x)=e^{2x}-a\ln x$.
(1)讨论$f(x)$的导函数${f}' (x)$零点的个数。
(2)证明：当$a\gt 0$时，$f(x)\ge 2a+a\ln \cfrac{2}{a} $.
$解(1)f(x)的定义域为(0,+\infty). \quad {f}' (x)=2e^{2x}-\cfrac{a}{x}(x\gt 0).\quad 当a\le 0时， {f}' (x)\gt 0,{f}' (x)没有零点。$
$当a\gt 0时，\because 2e^{2x}单调递增，-\cfrac{a}{x}单调递增；$
${\color{Green} \because } {f}' (a)=2e^{2a}-\cfrac{a}{a}=2e^{2a}\gt 0,$
${\color{Green} \therefore } 当0\lt b \lt \cfrac{a}{4} 且b\lt \cfrac{1}{4} 时，{f}' (b)\lt 0，即a\gt 0时{f}' (x)存在唯一零点。$
$(2)证明：设{f}' (x)在(0,+\infty)的唯一零点为x_0。当x\in (0,x_0)时，{f}' (x)\lt 0;当x\in (x_0,\infty)时，$
$故f(x)在(0,x_0)单调递减，在(x_0,+\infty)单调递增，f(x)在x=x_0有极小值f(x_0).$
${\color{Green} {f}' (x_0)} =2e^{2x_0}-\cfrac{a}{x_0}=0\Rightarrow 2e^{2x_0}=\cfrac{a}{x_0}$
$两边取对数，得{\color{Green} 2x_0=\ln \cfrac{a}{2x_0}}$
$f(x_0)=e^{2x_0}-a\ln x_0=\cfrac{a}{2x_0}-a\ln x_0=\cfrac{a}{2x_0}+2ax_0-2ax_0-a\ln x_0=\cfrac{a}{2x_0}+2ax_0-a({\color{Green} 2x_0} +\ln x_0)$
$=\cfrac{a}{2x_0}+2ax_0-a({\color{Green} \ln \cfrac{a}{2x_0}} +\ln x_0)=\cfrac{a}{2x_0}+2ax_0-a\ln \cfrac{a}{2}=\cfrac{a}{2x_0}+2ax_0+a\ln \cfrac{2}{a}\ge 2a+a\ln \cfrac{2}{a}$

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！！
$S_n=\tan 3n$