将函数\(f(x) = {e^x}\)展开成\(x\)的幂级数为（ ）。
A: \({e^x} = \sum\limits_{n = 0}^\infty { { { { x^n}} \over {n!}}} ( - \infty < x < + \infty )\)
B: \({e^x} = \sum\limits_{n = 0}^\infty { { {( - 1)}^n} { { {x^n}} \over {n!}}} ( - \infty < x < + \infty )\)
C: \({e^x} = \sum\limits_{n = 0}^\infty { { { { x^n}} \over {n!}}} ( - 1 < x < 1)\)
D: \({e^x} = \sum\limits_{n = 0}^\infty { { {( - 1)}^n} { { {x^n}} \over {n!}}} ( - 1 < x < 1)\)
A: \({e^x} = \sum\limits_{n = 0}^\infty { { { { x^n}} \over {n!}}} ( - \infty < x < + \infty )\)
B: \({e^x} = \sum\limits_{n = 0}^\infty { { {( - 1)}^n} { { {x^n}} \over {n!}}} ( - \infty < x < + \infty )\)
C: \({e^x} = \sum\limits_{n = 0}^\infty { { { { x^n}} \over {n!}}} ( - 1 < x < 1)\)
D: \({e^x} = \sum\limits_{n = 0}^\infty { { {( - 1)}^n} { { {x^n}} \over {n!}}} ( - 1 < x < 1)\)
举一反三
- 将\(f(x) = {1 \over {1 + {x^2}}}\)展开成\(x\)的幂级数为( )。 A: \({1 \over {1 + {x^2}}} = \sum\limits_{n = 0}^\infty { { {( - 1)}^n}{x^{2n}}} \matrix{ {} & {} \cr } ( - \infty < x < + \infty )\) B: \({1 \over {1 + {x^2}}} = \sum\limits_{n = 0}^\infty { { {( - 1)}^n}{x^{2n}}} \matrix{ {} & {} \cr } ( - 1< x < 1)\) C: \({1 \over {1 + {x^2}}} = \sum\limits_{n = 0}^\infty { { {( - 1)}^n}{x^{2n}}} \matrix{ {} & {} \cr } ( - 1 < x < 1)\) D: \({1 \over {1 + {x^2}}} = \sum\limits_{n = 0}^\infty { { x^{2n}}} \matrix{ {} & {} \cr } ( - 1 < x < 1)\)
- 将\(f(x)=e^x\)展开成\((x-3)\)的幂级数为( )。 A: \(\sum\limits_{n = 0}^\infty { { { { {(x - 3)}^n}} \over {n!}}} \matrix{ {} & {} \cr } ( - 1, 1)\) B: \({e^3}\sum\limits_{n = 0}^\infty { { { { {(x - 3)}^n}} \over {n!}}} \matrix{ {} & {} \cr } ( - 1, 1)\) C: \(\sum\limits_{n = 0}^\infty { { { { {(x - 3)}^n}} \over {n!}}} \matrix{ {} & {} \cr } ( - \infty , + \infty )\) D: \({e^3}\sum\limits_{n = 0}^\infty { { { { {(x - 3)}^n}} \over {n!}}} \matrix{ {} & {} \cr } ( - \infty , + \infty )\)
- 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: \({1 \over {2 - x}} = \sum\limits_{n = 0}^\infty { { { { x^n}} \over { { 2^{n + 1}}}}} \)，\(\left( { - 2,2} \right]\)
- 设幂级数\(\sum\limits_{n = 0}^\infty { { a_n}} {x^n}\)与\(\sum\limits_{n = 1}^\infty { { b_n}{x^n}} \)的收敛半径分别为\( { { \sqrt 5 } \over 3}\)与\({1 \over 3}\)，则幂级数\(\sum\limits_{n = 1}^\infty { { {a_n^2} \over {b_n^2}}} {x^n}\)的收敛半径为( )。 A: 5 B: \( { { \sqrt 5 } \over 3}\) C: \({1 \over 3}\) D: \({1 \over 5}\)
- 下面级数求和错误的是 A: $\sum_{n=0}^\infty q^n = \frac{1}{1-q} (0\lt q\lt1) $ B: $\sum_{n=1}^\infty \frac{x^{2^{n-1}}}{1-x^{2^n}} = \frac{x}{1-x} (|x|\lt 1) $ C: $\sum_{n=1}^\infty \frac{1}{{n!}} = e $ D: $\sum_{n=1}^\infty \frac{x^{2^{n-1}}}{1-x^{2^n}} = \frac{1}{1-x} (x>1) $