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            • 1.

              数列\(\{ a_{n}\}\)前\(n\)项和为\(S_{n}{=}n^{2}{+}3n+1\),则\(\{ a_{n}\}\)的通项等于______ .

            • 2.

              设\(S_{n}\)为正项数列\(\{ a_{n}\}\)的前\(n\)项和,\(a_{1}{=}2{,}S_{n{+}1}(S_{n{+}1}{-}2S_{n}{+}1){=}3S_{n}(S_{n}{+}1)\),则\(a_{100}\)等于\(({  })\)

              A.\(2{×}3^{98}\)
              B.\(4{×}3^{98}\)
              C.\(2{×}3^{99}\)
              D.\(4{×}3^{99}\)
            • 3. 数列{an}满足a1=2,
              (1)设,求数列{bn}的通项公式;
              (2)设,数列{cn}的前n项和为Sn,求出Sn并由此证明:
            • 4. 数列{an}满足Sn=2n-an(n∈N*).
              (Ⅰ)计算a1,a2,a3,a4,并由此猜想通项公式an
              (Ⅱ)用数学归纳法证明(Ⅰ)中的猜想.
            • 5.
              设数列\(\{a_{n}\}\)的前\(n\)项和为\(S_{n}\),已知\(a_{1}=1\),\(a_{n+1}= \dfrac {2n+3}{n}S_{n}(n∈N^{*}).\)
              \((1)\)证明:数列\(\{ \dfrac {S_{n}}{n}\}\)是等比数列;
              \((2)\)求数列\(\{S_{n}\}\)的前\(n\)项和\(T_{n}\).
            • 6.
              已知数列\(\{a_{n}\}\)满足\(a_{1}=1\),\(a_{n+1}+a_{n}= \sqrt {n+1}- \sqrt {n-1}\),\(n∈N^{*}\).
              \((\)Ⅰ\()\)求\(a_{2}\),\(a_{3}\),\(a_{4}\);
              \((\)Ⅱ\()\)猜想数列\(\{a_{n}\}\)的通项公式,并用数学归纳法证明.
            • 7. 已知数列\(\{ \)\(a_{n}\)\(\}\)中, \(a\)\({\,\!}_{2}=\) \(a\)\(( \)\(a\)为非零常数\()\),其前 \(n\)项和\(S\) \({\,\!}_{n}\)满足:\(S\) \({\,\!}_{n}\)\(= \dfrac{n\left({a}_{n}-{a}_{1}\right)}{2}\left(n∈{N}^{*}\right) \)
              \((1)\)求数列\(\{ \)\(a_{n}\)\(\}\)的通项公式;
              \((2)\)若 \(a\)\(=2\),且\( \dfrac{1}{4} \) \(a_{m}\)\({\,\!}^{2}-S\) \({\,\!}_{n}\)\(=11\),求 \(m\)\(n\)的值;
            • 8.

              已知函数\(f(x)=\dfrac{x}{\sqrt{1+{{x}^{2}}}}(x > 0)\),数列\(\left\{ {{a}_{n}} \right\}\)满足\({{a}_{1}}=f(x)\),\({{a}_{n+1}}=f({{a}_{n}})\)

              \((\)Ⅰ\()\)求\({{a}_{2}}\),\({{a}_{3}}\),\({{a}_{4}}\);

              \((\)Ⅱ\()\)猜想数列\(\left\{ {{a}_{n}} \right\}\)的通项,并予以证明。

            • 9.

              数列\(\{\)\(a_{n}\)\(\}\)的通项\({a}_{n}={n}^{2}\left({\cos }^{2} \dfrac{nπ}{3}-{\sin }^{2} \dfrac{nπ}{3}\right) \),其前\(n\)项和为\(S_{n\;}\)\({\,\!}_{。}\)

              \((1)\)求\(S_{1}\),\(S_{2\;}\),\(S_{3}\);

              \((2)\)求\(S_{n}\)\({\,\!}\)

              \((3)\)若数列\({b}_{n}=- \dfrac{9n-4}{n+2}· \dfrac{1}{{S}_{3n-1}} \),其前\(n\)项和为\(T_{n}\),求证:\( \dfrac{2}{3}\leqslant {T}_{n}\leqslant \dfrac{3}{2} \)

            • 10.

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

              \((1)\)设\({{b}_{n}}=\dfrac{{{a}_{n}}}{{{2}^{n}}}\),求数列\(\left\{ {{b}_{n}} \right\}\)的通项公式;  

              \((2)\)求数列\(\left\{ {{a}_{n}} \right\}\)的前\(n\)项和\({{S}_{n}}\);

              \((3)\)记\({{c}_{n}}=\dfrac{{{\left( -1 \right)}^{n}}\left( {{n}^{2}}+4n+2 \right){{2}^{n}}}{{{a}_{n}}{{a}_{n+1}}}\),求数列\(\left\{ {{c}_{n}} \right\}\)的前\(n\)项和\({{T}_{n}}\).

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