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            • 1.
              已知数列:\( \dfrac {1}{1}\),\( \dfrac {2}{1}\),\( \dfrac {1}{2}\),\( \dfrac {3}{1}\),\( \dfrac {2}{2}\),\( \dfrac {1}{3}\),\( \dfrac {4}{1}\),\( \dfrac {3}{2}\),\( \dfrac {2}{3}\),\( \dfrac {1}{4}\),\(…\),依它的前\(10\)项的规律,这个数列的第\(2018\)项\(a_{2018}\)等于\((\)  \()\)
              A.\( \dfrac {1}{31}\)
              B.\( \dfrac {1}{63}\)
              C.\(64\)
              D.\( \dfrac {63}{2}\)
            • 2.
              大衍数列,来源于\(《\)乾坤谱\(》\)中对易传“大衍之数五十”的推论\(.\)主要用于解释中国传统文化中的太极衍生原理\(.\)数列中的每一项,都代表太极衍生过程中,曾经经历过的两仪数量总和\(.\)是中华传统文化中隐藏着的世界数学史上第一道数列题\(.\)其前\(10\)项依次是\(0\)、\(2\)、\(4\)、\(8\)、\(12\)、\(18\)、\(24\)、\(32\)、\(40\)、\(50…\),则此数列第\(20\)项为\((\)  \()\)
              A.\(180\)
              B.\(200\)
              C.\(128\)
              D.\(162\)
            • 3.
              已知数列\(\{a_{n})\)的通项公式为\(a_{n}= \dfrac {1+(-1)^{n+1}}{2}\),则该数列的前\(4\)项依次为\((\)  \()\)
              A.\(1\),\(0\),\(1\),\(0\)
              B.\(0\),\(1\),\(0\),\(1\)
              C.\( \dfrac {1}{2},0, \dfrac {1}{2},0\)
              D.\(2\),\(0\),\(2\),\(0\)
            • 4. 数列\( \dfrac {1}{3}\),\( \dfrac {1}{8}\),\( \dfrac {1}{15}\),\( \dfrac {1}{24}\),\(…\)的一个通项公式为\((\)  \()\)
              A.\(a_{n}= \dfrac {1}{2^{n}+1}\)
              B.\(a_{n}= \dfrac {1}{n+2}\)
              C.\(a_{n}= \dfrac {1}{n(n+2)}\)
              D.\(a_{n}= \dfrac {1}{2^{n}-1}\)
            • 5.

              设\({S}_{n} \)是数列\(\left\{{a}_{n}\right\} \)的前\(n\)项和,且\({a}_{1}=1,{a}_{n+1}=-{S}_{n}{S}_{n+1} \),则使\(\dfrac{nS_{n}^{2}}{1+10S_{n}^{2}} \)取得最大值时\(n\)的值为     \((\)    \()\)

              A.\(2\)              
              B.\(3\)                
              C.\(4\)               
              D.\(5\)
            • 6. 数列\( \dfrac {1}{2},- \dfrac {1}{4}, \dfrac {1}{8},- \dfrac {1}{16},…\)的一个通项公式可能是\((\)  \()\)
              A.\((-1)^{n} \dfrac {1}{2n}\)
              B.\((-1)^{n} \dfrac {1}{2^{n}}\)
              C.\((-1)^{n-1} \dfrac {1}{2n}\)
              D.\((-1)\;^{n-1} \dfrac {1}{2^{n}}\)
            • 7.
              数列\(6\),\(9\),\(14\),\(21\),\(30\),\(…\)的一个通项公式是\((\)  \()\)
              A.\(3n+3\)
              B.\(2n^{2}+1\)
              C.\(2^{n}+n+3\)
              D.\(n^{2}+5\)
            • 8.

              数列\(\sqrt{2}{,}\sqrt{5}{,}2\sqrt{2}{,}\sqrt{11}{…}\),的一个通项公式是\(({  })\)

              A.\(a_{n}{=}\sqrt{3n{-}3}\)
              B.\(a_{n}{=}\sqrt{3n{-}1}\)   
              C.\(a_{n}{=}\sqrt{3n{+}1}\)
              D.\(a_{n}{=}\sqrt{3n{+}3}\)
            • 9.
              已知数列\(\{a_{n}\}\)对任意的\(p\),\(q∈N^{*}\)满足\(a_{p+q}=a_{p}+a_{q}\),且\(a_{2}=—6\),那么\(a_{10}\)等于\((\)     \()\)
              A.\(—165\)
              B.\(—33\)
              C.\(—30\)
              D.\(—21\)
            • 10.

              数列\(1{,}3{,}5{,}7{,}9{,}{…}\)的通项公式为\((\ {  }\ )\)

              A.\(a_{n}{=}1{-}2n\)
              B.\(a_{n}{=}2n{-}1\)
              C.\(a_{n}{=}3n{-}1\)
              D.\(a_{n}{=}2n{+}1\)
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