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: \({1 \over {1 + {x^2}}} = \sum\limits_{n = 0}^\infty { { x^{2n}}} \matrix{ {} & {} \cr } ( - 1 < x < 1)\)

\( \lim \limits_{n \to \infty } { { n!} \over { { n^n}}} = \)______ 。

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

${X_1},{X_2},...,{X_n}$是来自均匀分布X~U(-a,a)的样本，用矩估计法估计参数a为() A: ${(\frac{3}{n}\sum\limits_{k = 1}^n {x_k^2} )^{\frac{1}{2}}}$ B: ${(\frac{2}{n}\sum\limits_{k = 1}^n {x_k^2} )^{\frac{1}{2}}}$ C: ${(\frac{3}{n}\sum\limits_{k = 1}^n {x_k} )^{\frac{1}{2}}}$ D: ${(\frac{2}{n}\sum\limits_{k = 1}^n {x_k} )^{\frac{1}{2}}}$

(2). 根据最小二乘法的思想，拟合直线回归方程是使（ ）。 A: \( \min\sum\limits_{i=1}^n {\vert y_i -\hat {y}\vert }<br/>\) B: \( \min\sum\limits_{i=1}^n {(y_i -\bar {y})^2} \) C: \( \min\sum\limits_{i=1}^n {(\hat {y}_i -\bar {y})^2}<br/>\) D: \( \min\sum\limits_{i=1}^n {(y_i -\hat {y}_i )^2} \)