优优班--学霸训练营 > 知识点挑题
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            • 1. 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.
            • 2. 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\( \dfrac {a_{m}}{a_{n}}∈( \dfrac {1}{6},6)\).
            • 3.

              已知数列\(\left\{ {{a}_{n}} \right\}\)满足:\({{a}_{1}}=1\),\({{a}_{n+1}}=\dfrac{{{a}_{n}}}{{{a}_{n}}+2}\) \(\left( n\in {{N}^{*}} \right).\)若\({{b}_{n+1}}=\left( n-2\lambda \right)\cdot \left( \dfrac{1}{{{a}_{n}}}+1 \right)\) \(\left( n\in {{N}^{*}} \right)\),\({{b}_{1}}=-\lambda \),且数列\(\left\{ {{b}_{n}} \right\}\)是单调递增数列,则实数\(\lambda \)的取值范围是____。

              A.\(\lambda > \dfrac{2}{3}\)
              B.\(\lambda > \dfrac{3}{2}\)
              C.\(\lambda < \dfrac{2}{3}\)
              D.\(\lambda < \dfrac{3}{2}\)
            • 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\(\{a_{n}\}\)的首项\(a_{1}=1\),\(a_{n+1}=3S_{n}(n\geqslant 1)\),则数列\(\{a_{n}\}\)的通项公式为 ______ .
            • 6. 数列\(\{a_{n}\}\)中,\(a_{1}= \dfrac {1}{2}\),且\((n+2)a_{n+1}=na_{n}\),则它的前\(20\)项之和\(S_{20}=(\)  \()\)
              A.\( \dfrac {18}{19}\)
              B.\( \dfrac {19}{20}\)
              C.\( \dfrac {20}{21}\)
              D.\( \dfrac {21}{22}\)
            • 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. 若数列\(\{a_{n}\}\)满足\(a_{n}-(-1)^{n}a_{n-1}=n(n\geqslant 2)\),\(S_{n}\)是\(\{a_{n}\}\)的前\(n\)项和,则\(S_{40}=\)________.
            • 9.

              \(S_{n}\)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\(T_{n}\)
            • 10. 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.

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