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            • 1.

              数列\(\{a_{n}\}\)满足\(a_{1}=1\),\(a_{n+1}=(n^{2}+n-λ)a_{n}(n=1,2,…)\),\(λ\)是常数.

              \((1)\)当\(a_{2}=-1\)时,求\(λ\)及\(a_{3}\)的值;

              \((2)\)是否存在实数\(λ\)使数列\(\{a_{n}\}\)为等差数列?若存在,求出\(λ\)及数列\(\{a_{n}\}\)的通项公式;若不存在,请说明理由.

            • 2. 已知函数\(f(x)= \dfrac {3}{2}x+\ln (x-1)\),设数列\(\{a_{n}\}\)同时满足下列两个条件:\(①a_{n} > 0(n∈N^{*})\);\(②a_{n+1}=f′(a_{n}+1)\).
              \((\)Ⅰ\()\)试用\(a_{n}\)表示\(a_{n+1}\);
              \((\)Ⅱ\()\)记\(b_{n}=a_{2n}(n∈N^{*})\),若数列\(\{b_{n}\}\)是递减数列,求\(a_{1}\)的取值范围.
            • 3. 已知数列\(\{a_{n}\}\)与\(\{b_{n}\}\)满足\(a_{n+1}-a_{n}=2(b_{n+1}-b_{n})\),\(n∈N^{*}\).
              \((1)\)若\(b_{n}=3n+5\),且\(a_{1}=1\),求\(\{a_{n}\}\)的通项公式;
              \((2)\)设\(\{a_{n}\}\)的第\(n_{0}\)项是最大项,即\(a_{n\_{0}}\geqslant a_{n}(n∈N*)\),求证:\(\{b_{n}\}\)的第\(n_{0}\)项是最大项;
              \((3)\)设\(a_{1}=3λ < 0\),\(b_{n}=λ^{n}(n∈N^{*})\),求\(λ\)的取值范围,使得对任意\(m\),\(n∈N^{*}\),\(a_{n}\neq 0\),且\( \dfrac {a_{m}}{a_{n}}∈( \dfrac {1}{6},6)\).
            • 4.

              设正项数列\(\left\{{a}_{n}\right\} \)的前\(n\)项和为\({S}_{n} \),且满足\({S}_{n}= \dfrac{1}{2}{{a}_{n}}^{2}+ \dfrac{n}{2}\left(n∈N*\right) \).

              \((1)\)计算\({a}_{1}\;,\;{a}_{2\;},\;{a}_{3} \)的值,并猜想\(\left\{{a}_{n}\right\} \)的通项公式;

              \((2)\)用数学归纳法证明\(\left\{{a}_{n}\right\} \)的通项公式.

            • 5. 已知数列\(\{ a_{n}\}\)的前\(n\)项和为\(S_{n}\),且\(S_{n}{=}a_{n}{+}n^{2}{-}1(n{∈}N^{{*}})\).
              \((1)\)求数列\(\{ a_{n}\}\)的通项公式;
              \((2)\)定义\(x{=[}x{]+ < }x{ > }\),其中\({[}x{]}\)为实数\(x\)的整数部分,\({ < }x{ > }\)为\(x\)的小数部分,且\(0{\leqslant < }x{ > < }1\),记\(c_{n}{= < }\dfrac{a_{n}a_{n{+}1}}{S_{n}}{ > }\),求数列\(\{ c_{n}\}\)的前\(n\)项和\(T_{n}\).
            • 6. 已知数列\(\{a_{n}\}\)的前\(n\)项和为\(S_{n}\),且满足\(S_{n}=2a_{n}-2\).
              \((\)Ⅰ\()\)求数列\(\{a_{n}\}\)的通项公式;
              \((\)Ⅱ\()\)设函数\(f(x)=( \dfrac {1}{2})^{x}\),数列\(\{b_{n}\}\)满足条件\(b_{1}=2\),\(f(b_{n+1})= \dfrac {1}{f(-3-b_{n})}\),\((n∈N^{*})\),若\(c_{n}= \dfrac {b_{n}}{a_{n}}\),求数列\(\{c_{n}\}\)的前\(n\)项和\(T_{n}\).
            • 7. 已知各项均为正数的数列\(\{a_{n}\}\)的前\(n\)项和\(S_{n}\)满足\(S_{n} > 1\)且\(6S_{n}=(a_{n}+1)(a_{n}+2)\),\(n∈N^{*}\)
              \((1)\)求数列\(\{a_{n}\}\)的通项公式;
              \((2)\)若数列\(\{b_{n}\}\)的前\(n\)项的和为\(b_{n}=-a_{n}+19\),求数列\(\{|b_{n}|\}\)的前\(n\)项和\(T_{n}\).
            • 8.

              \(S_{n}\)是数列\(\{ a_{n}\}\)的前\(n\)项和,已知\({a}_{1}=1,{a}_{n+1}=2{S}_{n}+1(n∈{N}^{∗}) \)

              \((\)Ⅰ\()\)求数列\(\{ a_{n}\}\)的通项公式;
              \((\)Ⅱ\()\)\(\dfrac{b_{n}}{a_{n}}{=}3n{-}1\),求数列\(\{ b_{n}\}\)的前\(n\)项和\(T_{n}\)
            • 9. 若数列\(\{a_{n}\}\)的前\(n\)项和为\(S_{n}\),且满足\(a_{n}+2S_{n}S_{n-1}=0(n\geqslant 2)\),\(a_{1}= \dfrac{1}{2}\).
              \((1)\)求证:\(\left\{ \left. \dfrac{1}{S_{n}} \right. \right\}\)成等差数列;

              \((2)\)求数列\(\{a\)\({\,\!}_{n}\)\(\}\)的通项公式.

            • 10.

              单调递增数列\(\left\{ {{a}_{n}} \right\}\)的前\(n\)项和为\({{S}_{n}}\),且满足\(4{{S}_{n}}=a_{n}^{2}+4n\).

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