• 2022-06-06
    将\(f(x) = {1 \over {2 - x}}\)展开成\(x \)的幂级数为( )。
    A: \({1 \over {2 - x}} = \sum\limits_{n = 0}^\infty { { { { x^n}} \over { { 2^{n }}}}} \),\(( - 2,2)\)
    B: \({1 \over {2 - x}} = \sum\limits_{n = 0}^\infty { { { { x^n}} \over { { 2^{n }}}}} \),\(\left( { - 2,2} \right]\)
    C: \({1 \over {2 - x}} = \sum\limits_{n = 0}^\infty { { { { x^n}} \over { { 2^{n + 1}}}}} \),\(( - 2,2)\)
    D: \({1 \over {2 - x}} = \sum\limits_{n = 0}^\infty { { { { x^n}} \over { { 2^{n + 1}}}}} \),\(\left( { - 2,2} \right]\)
  • C

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    • 0

      \( {1 \over {1 + x}} \)的麦克劳林公式为( )。 A: \( {1 \over {1 + x}} = 1 + x + { { {x^2}} \over 2} + \cdots + { { {x^n}} \over {n!}} + o\left( { { x^n}} \right) \) B: \( {1 \over {1 + x}} = 1 + x + {x^2} + \cdots + {x^n} + o\left( { { x^n}} \right) \) C: \( {1 \over {1 + x}} = 1 - x + {x^2} - \cdots + {( - 1)^n}{x^n} + o\left( { { x^n}} \right) \) D: \( {1 \over {1 + x}} = 1 - x - { { {x^2}} \over 2}- \cdots - { { {x^n}} \over {n!}} + o\left( { { x^n}} \right) \)

    • 1

      \( {1 \over {1 + x}} \)的麦克劳林公式为( ). A: \( {1 \over {1 + x}} = 1 + x + { { {x^2}} \over 2} + \cdots + { { {x^n}} \over {n!}} + o\left( { { x^n}} \right) \) B: \( {1 \over {1 + x}} = 1 + x + {x^2} + \cdots + {x^n} + o\left( { { x^n}} \right) \) C: \( {1 \over {1 + x}} = 1 - x + {x^2} - \cdots + {( - 1)^n}{x^n} + o\left( { { x^n}} \right) \)

    • 2

      \( \sin x \)的麦克劳林公式为( ). A: \( \sin x = x - { { {x^3}} \over {3!}} + { { {x^5}} \over {5!}} - \cdots + {( - 1)^n} { { {x^{2n + 1}}} \over {\left( {2n + 1} \right)!}} + o\left( { { x^{2n + 2}}} \right) \) B: \( \sin x = 1 - { { {x^2}} \over {2!}} + { { {x^4}} \over {4!}} - { { {x^6}} \over {6!}} + \cdots + {( - 1)^n} { { {x^{2n}}} \over {\left( {2n} \right)!}} + o\left( { { x^{2n + 1}}} \right) \) C: \( \sin x = 1 + x + { { {x^2}} \over 2} + \cdots + { { {x^n}} \over {n!}} + o\left( { { x^n}} \right) \)

    • 3

      \( \lim \limits_{x \to 0} { { \sqrt {1 + x\sin x} - \cos x} \over { { {\sin }^2}{x \over 2}}} = \)______ 。

    • 4

      \(\lim \limits_{n \to \infty } { { {\rm{3}}{n^2}{\rm{ + 8}}} \over { { n^2} - n}} = \) .______