优优班--学霸训练营 > 知识点挑题
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            • 1.

              数列\(\{a_{n}\}\)满足\(a_{1}=1\),且\(a_{n+1}=a_{1}+a_{n}+n(n∈N^{*})\),则\( \dfrac{1}{a_{1}}+ \dfrac{1}{a_{2}}+…+ \dfrac{1}{a_{2 016}}\)等于\((\)  \()\)

              A.\( \dfrac{4 032}{2 017}\)
              B.\( \dfrac{4 028}{2 015}\)

              C.\( \dfrac{2 015}{2 016}\)
              D.\( \dfrac{2 014}{2 015}\)
            • 2.

              当\(n\geqslant 2\)时,\( \dfrac{1}{n^{2}-1}= \dfrac{1}{2}\left( \left. \dfrac{1}{n-1}- \dfrac{1}{n+1} \right. \right).(\)  \()\)

              A.\(√\)  
              B.\(×\)
            • 3.

              \({{a}_{n}}=2{{n}^{2}}-n\),以下四个数是数列\(\left\{ {{a}_{n}} \right\}\)中的一项的是(    )

              A.\(30\)
              B.\(44\)
              C.\(66\)
              D.\(90\)
            • 4.

              数列\(-1,\dfrac{4}{3},-\dfrac{9}{5},\dfrac{16}{7},\cdots \)的一个通项公式是\((\)  \()\)

              A.\({{a}_{n}}={{\left( -1 \right)}^{n}}\dfrac{{{n}^{2}}}{2n-1}\)
              B.\({{a}_{n}}={{\left( -1 \right)}^{n}}\dfrac{n+1}{2n-1}\)
              C.\({{a}_{n}}={{\left( -1 \right)}^{n}}\dfrac{{{n}^{2}}}{2n+1}\)
              D.\({{a}_{n}}={{\left( -1 \right)}^{n}}\dfrac{{{n}^{2}}-2n}{2n-1}\)
            • 5.

              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.\(\lambda < \dfrac{3}{2}\)
            • 6. 已知数列\(\{{{a}_{n}}\}\)的前\(n\)项和为\({{S}_{n}}\),若\(3{{S}_{n}}=2{{a}_{n}}-3n\),则\({{a}_{2018}}=\)

              A.\({{2}^{2018}}-1\)
              B.\({{3}^{2018}}-6\)
              C.\({{\left( \dfrac{1}{2} \right)}^{2018}}-\dfrac{7}{2}\)
              D.\({{\left( \dfrac{1}{3} \right)}^{2018}}-\dfrac{10}{3}\) 
            • 7.

              已知数列\(\{a_{n}\}\)是首项为\(1\),公差为\(2\)的等差数列,数列\(\{b_{n}\}\)满足\(\dfrac{{{a}_{{1}}}}{{{b}_{{1}}}}+\dfrac{{{a}_{{2}}}}{{{b}_{{2}}}}+\dfrac{{{a}_{{3}}}}{{{b}_{{3}}}}+\ldots +\dfrac{{{a}_{n}}}{{{b}_{n}}}=\dfrac{{1}}{{{{2}}^{n}}}\),若数列\(\{b_{n}\}\)的前\(n\)项和为\(S_{n}\),则\(S_{5}=\)


              A.\(-454\)
              B.\(-450\)
              C.\(-446\)
              D.\(-442\)
            • 8.

              在数列\(\{{{a}_{n}}\}\)中,\({{a}_{\ 1}}=2,{{a}_{n+1}}={{a}_{n}}+\ln (1+\dfrac{1}{n})\),则\({{a}_{n}}=\)                     \((\)   \()\)

              A.\(2+(n-1)\ln n\)
              B.\(2+\ln n\)         
              C. \(2+n\ln n\)
              D.\(1+n+\ln n\)
            • 9. 已知数列\(\{ \)\(a_{n}\)\(\}\)的通项公式为 \(a_{n}\)\(= \dfrac{4}{11-2n}( \)\(n\)\(∈N^{*})\),则满足 \(a_{n}\)\({\,\!}_{+1} < \) \(a_{n}\)\(n\)的取值为(    )
              A.\(3\)       
              B.\(4\)
              C.\(5\)                                       
              D.\(6\)
            • 10.

              已知\(f\left(n\right)= \dfrac{1}{n+1}+ \dfrac{1}{n+2}+ \dfrac{1}{n+3}+...+ \dfrac{1}{2n}\left(n∈{N}^{*}\right), \)那么\(f\left(n+1\right)-f\left(n\right) \)等于   \((\)   \()\)

              A.\(\dfrac{1}{2n+1}\)
              B.\(\dfrac{1}{2n+2}\)
              C.\(\dfrac{1}{2n+1}+\dfrac{1}{2n+2}\)
              D.\(\dfrac{1}{2n+1}-\dfrac{1}{2n+2}\)
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