已知数列\(\{a_{n}\}\)满足\(n{a}_{n}-\left(n+1\right){a}_{n-1}=2{n}^{2}+2n(n=2,3,4...),{a}_{1}=6 \)

\((1)\)求证\(\left\{ \dfrac{{a}_{n}}{n+1}\right\} \)为等差数列，并求出\(\{a\)\(n\)\(\}\)的通项公式

\((2)\)数列\(\left\{ \dfrac{1}{{a}_{n}}\right\} \)的前\(n\)项和\(S_{n,}\)求求证：\({S}_{n} < \dfrac{5}{12} \)

已知数列\(\{{{a}_{n}}\}\)各项均为正数，\({{S}_{n}}\)为其前\(n\)项和，满足\({{S}_{n}}={{(\dfrac{1+{{a}_{n}}}{2})}^{2}}\)，

\((1)\)求\(\{{{a}_{n}}\}\)的通项公式；

\((2)\)令\({{b}_{n}}=a_{n}^{2},n\in {{N}^{*}}\)，证明：\(\dfrac{1}{{{b}_{1}}}+\dfrac{1}{{{b}_{2}}}+\dfrac{1}{{{b}_{3}}}+\cdots +\dfrac{1}{{{b}_{n}}} < \dfrac{5}{4}\)

已知数列\(\{a_{n}\}\)各项均为正数，且\(a_{1}=1\)，\(a_{n+1}a_{n}+a_{n+1}-a_{n}=0(n∈N^{*}).\)

\((1)\)设\(b_{n}= \dfrac{1}{a_{n}}\)，求证：数列\(\{b_{n}\}\)是等差数列；

\((2)\)求数列\(\left\{ \left. \dfrac{a_{n}}{n+1} \right. \right\}\)的前\(n\)项和\(S_{n}\)．

已知数列\(\left\{ a_{n} \right\}\)是公比为\(2\)的等比数列，且\(a_{2}\)，\(a_{3}{+}1\)，\(a_{4}\)成等差数列．

\((1)\)求数列\(\left\{ a_{n} \right\}\)的通项公式；

\((2)\)记\({{b}_{n}}={{\log }_{2}}{{a}_{n+1}}\)，\({{c}_{n}}=\dfrac{1}{{{b}_{n}}\cdot {{b}_{n+1}}}\)求数列\(\left\{ {{c}_{n}} \right\}\)的前\(n\)项和\(T_{n}\)．

设\({{S}_{n}}\)是等差数列\(\left\{ {{a}_{n}} \right\}\)的前\(n\)项和，若\({{a}_{1}}=-2015\)，\({{S}_{6}}-2{{S}_{3}}=18\)，则\({{S}_{2017}}=(\)  \()\)

三个正数\(a\)、\(b\)、\(c\)成等比数列，则\(\lg a\)、 \(\lg b\)、 \(\lg c\)是 (    )

在等比数列\(\left\{ {{a}_{n}} \right\}\)中，\(3{{a}_{1}},\dfrac{1}{2}{{a}_{5}},2{{a}_{3}}\)成等差数列，则\(\dfrac{{{a}_{9}}+{{a}_{10}}}{{{a}_{7}}+{{a}_{8}}}=\)              ．