Which one of the following sequences is not covergent?
A: un=∑nk=1sink2k,n=1,2,⋯.
B: un=cos(1!)1⋅2+cos(2!)2⋅3+cos(3!)3⋅4+⋯+cos(n!)n⋅(n+1),n=1,2,⋯.
C: un=∑nk=1(−1)k−11k,n=1,2,⋯.
D: un=(1+3n(−1)n)1/n,n=1,2,⋯.
A: un=∑nk=1sink2k,n=1,2,⋯.
B: un=cos(1!)1⋅2+cos(2!)2⋅3+cos(3!)3⋅4+⋯+cos(n!)n⋅(n+1),n=1,2,⋯.
C: un=∑nk=1(−1)k−11k,n=1,2,⋯.
D: un=(1+3n(−1)n)1/n,n=1,2,⋯.
举一反三
- 设`\n`阶方阵`\A`满足`\|A| = 2`，则`\|A^TA| = ,|A^{ - 1}| = ,| A^ ** | = ,| (A^ ** )^ ** | = ,|(A^ ** )^{ - 1} + A| = ,| A^{ - 1}(A^ ** + A^{ - 1})A| = `分别等于（ ） A: \[4,\frac{1}{2},{2^{n - 1}},{2^{{{(n - 1)}^2}}},2{(\frac{3}{2})^n},\frac{{{3^n}}}{2}\] B: \[2,\frac{1}{2},{2^{n - 1}},{2^{{{(n + 1)}^2}}},2{(\frac{3}{2})^n},\frac{{{3^n}}}{2}\] C: \[4,\frac{1}{2},{2^{n + 1}},{2^{{{(n - 1)}^2}}},2{(\frac{3}{2})^{n - 1}},\frac{{{3^n}}}{2}\] D: \[2,\frac{1}{2},{2^{n - 1}},{2^{{{(n - 1)}^2}}},2{(\frac{3}{2})^{n - 1}},\frac{{{3^n}}}{2}\]
- 函数\(f(x) = x^2,\; x \in [-\pi,\pi]\)的Fourier级数为 A: \(\frac{\pi^2}{3}+4\Sigma_{n=1}^{\infty} \frac{(-1)^n}{n^2} \sin nx ,\; x \in [-\pi,\pi]\) B: \(\frac{\pi^2}{3}+4\Sigma_{n=1}^{\infty} \frac{(-1)^n}{n^2} \cos nx ,\; x \in [-\pi,\pi]\) C: \(\frac{2\pi^2}{3}+4\Sigma_{n=1}^{\infty} \frac{(-1)^n}{n^2} \sin nx ,\; x \in [-\pi,\pi]\) D: \(\frac{2\pi^2}{3}+4\Sigma_{n=1}^{\infty} \frac{(-1)^n}{n^2} \cos nx ,\; x \in [-\pi,\pi]\)
- 设随机变量X的分布律为P{X=k}=a/N，k=1,2,…,N，则a=( ) A: 2 B: 1 C: 1/2 D: 1/3
- 下面程序的功能是输出以下9阶方阵。请填空。 1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 1 1 2 3 3 3 3 3 2 1 1 2 3 4 4 4 3 2 1 1 2 3 4 5 4 3 2 1 1 2 3 4 4 4 3 2 1 1 2 3 3 3 3 3 2 1 1 2 2 2 2 2 2 2 1 1 1 1 1 1 1 1 1 1 # include int main( ) { int a[10][10]，n，i，j，m； scanf("%d"，&n)； if(n%2= =0) m=n/2； else( )； for(i=0；i m=n/2+1 n–i–1 n–i–1
- ${X_1},{X_2},...,{X_n}$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: ${(\frac{2}{n}\sum\limits_{k = 1}^n {x_k} )^{\frac{1}{2}}}$