已知数列\(\left\{ {{a}_{n}} \right\}\)中，\({a}_{1}=2,{a}_{n+1}= \dfrac{4{a}_{n}-2}{3{a}_{n}-1}\left(n∈{N}^{*}\right) \)，设\({{b}_{n}}=\dfrac{3{{a}_{n}}-2}{{{a}_{n}}-1}\)．

\((\)Ⅰ\()\)试写出数列\(\left\{ {{b}_{n}} \right\}\)的前三项；

\((\)Ⅱ\()\)求证：数列\(\left\{ {{b}_{n}} \right\}\)是等比数列，并求数列\(\left\{ {{a}_{n}} \right\}\)的通项公式\({{a}_{n}}\)；

\((\)Ⅲ\()\)设\(\left\{ {{a}_{n}} \right\}\)的前\(n\)项和为\({{S}_{n}}\)，求证：\(\dfrac{\left( n+1 \right)\cdot {{2}^{n+1}}-n-2}{{{2}^{n+1}}-1} < {{S}_{n}}\leqslant \dfrac{\left( n+2 \right)\cdot {{2}^{n-1}}-1}{{{2}^{n-1}}}\left( n\in N* \right)\)．

已知\(α∈\left(0, \dfrac{π}{2}\right) \)，\(β∈\left(0, \dfrac{π}{2}\right) \)，且\(\dfrac{\alpha }{2(1+\cos \dfrac{\alpha }{2})} < \tan \beta < \dfrac{1-\cos \alpha }{\alpha }\)，则

设函数\(f(x)= \dfrac{a{x}^{2}+bx+1}{x+c}(a > 0) \)为奇函数，且\(|f(x){|}_{min}=2 \sqrt{2} \)，数列\(\{a\)\({\,\!}_{n}\)\(\}\)与\(\{b\)\({\,\!}_{n}\)\(\}\)满足如下关系：\({a}_{1}=2,{a}_{n+1}= \dfrac{f({a}_{n})-{a}_{n}}{2},{b}_{n}= \dfrac{{a}_{n}-1}{{a}_{n}+1} \)．

\((1)\)求\(f(x)\)的解析式；

\((3)\)记\({{S}_{n}}\)为数列\(\{{{a}_{n}}\}\)的前\(n\)项和，求证：对任意\(n\in {{N}^{*}}\)有\({{S}_{n}} < n+\dfrac{3}{2}.\)

已知函数\(f\left( x \right)=\dfrac{{lo}{{{g}}_{3}}\left( x+1 \right)}{x+1}\left( x > 0 \right)\)的图象上有一点列\({{P}_{n}}\left( {{x}_{n}},{{y}_{n}} \right)\left( n\in {{N}^{*}} \right)\)，点\({{P}_{n}}\)在\(x\)轴上的射影是\({{Q}_{n}}\left( {{x}_{n}},0 \right)\)，且\({{x}_{n}}=3{{x}_{n-1}}+2\) \((n\geqslant 2\)且\(n\in {{N}^{*}})\)，\({{x}_{1}}=2\)．

\((1)\)求证：\(\left\{ {{x}_{n}}+1 \right\}\)是等比数列，并求出数列\(\left\{ {{x}_{n}} \right\}\)的通项公式；

\((2)\)对任意的正整数\(n\)，当\(m\in \left[ -1,1 \right]\)时，不等式\(3{{t}^{2}}-6mt+\dfrac{1}{3} > {{y}_{n}}\)恒成立，求实数\(t\)的取值范围．

\((3)\)设四边形\({{P}_{n}}{{Q}_{n}}{{Q}_{n+1}}{{P}_{n+1}}\)的面积是\({{S}_{n}}\)，求证：\(\dfrac{1}{{{S}_{1}}}+\dfrac{1}{2{{S}_{2}}}+\ldots +\dfrac{1}{n{{S}_{n}}} < 3\)．

下列结论不正确的是(    )

\(①.\dfrac{1}{{{2}^{10}}}+\dfrac{1}{{{2}^{10}}+1}+\dfrac{1}{{{2}^{10}}+2}+\cdots +\dfrac{1}{{{2}^{11}}-1} > 1\)

\(②.\)若\(|a| < 1\)，则\(|a+b|-|a-b|\ > 2\)

\(③.\lg 9\cdot \lg 11 < 1\)

\(④.\)若\(x > 0,y > 0\)，则\(\dfrac{x+y}{1+x+y} < \dfrac{x}{1+x}+\dfrac{y}{1+y}\)