对于给定非负整数[img=11x14]1803c4615c4a826.png[/img]，计算[img=132x24]1803c461653d41a.png[/img]的方法是
A: [x^2 | x - [1..n]]
B: sum[x^2 | x - [1..n]]
C: sum [x | x^2 - [1..n]]
D: sum [x | x - [1..n^2]]
E: sum [y | y- [1..n], y = x^2]
F: sum [ x*y | (x,y) - zip [1..n] [1..n]]
G: sum [ x*x | (x,x) - zip [1..n] [1..n]]
A: [x^2 | x - [1..n]]
B: sum[x^2 | x - [1..n]]
C: sum [x | x^2 - [1..n]]
D: sum [x | x - [1..n^2]]
E: sum [y | y- [1..n], y = x^2]
F: sum [ x*y | (x,y) - zip [1..n] [1..n]]
G: sum [ x*x | (x,x) - zip [1..n] [1..n]]
举一反三
- 对于给定非负整数[img=11x14]17de86c7a4e3414.png[/img]，计算[img=132x24]17de86c7b1aebe2.png[/img]的方法是 A: [x^2 | x - [1..n]] B: sum[x^2 | x - [1..n]] C: sum [x | x^2 - [1..n]] D: sum [x | x - [1..n^2]] E: sum [y | y- [1..n], y = x^2] F: sum [ x*y | (x,y) - zip [1..n] [1..n]] G: sum [ x*x | (x,x) - zip [1..n] [1..n]]
- 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: \({1 \over {2 - x}} = \sum\limits_{n = 0}^\infty { { { { x^n}} \over { { 2^{n + 1}}}}} \)，\(\left( { - 2,2} \right]\)
- 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: \({e^x} = \sum\limits_{n = 0}^\infty { { {( - 1)}^n} { { {x^n}} \over {n!}}} ( - 1 < x < 1)\)
- 已知()y()=()ln()x(),则()y()(()n())()=()。A.()(()−()1())()n()n()!()x()−()n()"()role="presentation">()(()−()1())()n()n()!()x()−()n();()B.()(()−()1())()n()(()n()−()1())()!()x()−()2()n()"()role="presentation">()(()−()1())()n()(()n()−()1())()!()x()−()2()n();()C.()(()−()1())()n()−()1()(()n()−()1())()!()x()n()"()role="presentation">()(()−()1())()n()−()1()(()n()−()1())()!()x()-n();()D.()(()−()1())()n()−()1()n()!()x()−()n()+()1()"()role="presentation">()(()−()1())()n()−()1()n()!()x()−()n()+()1().