解析 13.答案:$$ \frac { m ^ { 9 } } { 1 2 n } $$ 解析:根据运算性质得,9个m相乘是 $$ m ^ { 9 } $$,12个n相加是12n,故原式= $$ \frac { m ^ { 9 } } { 1 2 n } $$.故答案为$$ \frac { m ^ { 9 } } { 1 2 n } $$ ...
结果1 题目 $$ 1 \cdot 2 ^ { 2 } + 2 \cdot 3 ^ { 2 } + \cdots n ( n + 1 ) ^ { 2 } = \frac { 1 } { 1 2 } n ( n + 1 ) ( a n ^ { 2 } + b n + 1 0 ) $$$ a = \_ , b = \_ . $$ 相关知识点: 整式加减 规律探究与定义...
因为Y1~Yn服从标准正态分布,而Zi为其线性组合,因此Zi也服从正态分布。 E(Zi)=E(ai1Y1+ai2Y2+ai3Y3+……+ainYn)=∑k=1naikE(Yk) 由于E(Yk)=0,所以:E(Zi)=0; D(Zi)=ai12D(Y1)+ai22D(Y2)+ai32D(Y3)+……+ain2D(Yn)=∑k=1naik2D(Y) 由(1)式得:D(Z_{i})=1\cdot D(Y)=1。
\partial {{a}_{j}}}+\sum\limits_{i,j,k=1}^{2}{\frac{\Delta {{x}_{i}}\Delta {{x}_{j}}\Delta {{x}_{k}}}{3!}\frac{{{\partial }^{3}}f\left( {\vec{a}} \right)}{\partial {{a}_{i}}\partial {{a}_{j}}\partial {{a}_{k}}}+\cdot \cdot \cdot }\] ...
\frac{g_2^2}{4\,m_w^2}\,\mathfrak {r}\mathfrak {e}{\left[ x_{id}\,x_{jd}^* + x_{iu}\,x_{ju}^*\right] } \nonumber \\&\qquad \left( 2\,p^2 - m_{h_i}^2 - m_{h_j}^2+\tilde{c}_a\right) a_{0}{\left( \xi \, m_w^2\right) }\,. \end{...
结果1 题目 6. 计算$$ \frac { \frac { 9 个 } { m \cdot m \cdot \cdots \cdot m } } { \frac { n + n + \cdots + n } { 1 2 0 } } = \_ . $$ 相关知识点: 试题来源: 解析 6.$$ \frac { m ^ { 9 } } { 1 2 n } $$ 反馈 收藏 ...
结果1 题目 计算$$ \frac { ( 2 ^ { n + 1 } ) ^ { 2 } \cdot ( \frac { 1 } { 2 } ) ^ { 2 n + 1 } } { 4 ^ { n } \cdot 8 ^ { - 2 } } ( n \in N _ { + } $$的结果为 ( ) A.$$ \frac { 1 } { 6 4 } $$ B.$$ 2 ^ { 2...
\begin{aligned} 2\pi Q_k(n) &:=\frac1i\int_{-\infty}^{(0+)}u^{1/2}\exp\left({\pi\over12u}+{2\pi u\lambda_n^2\over k^2}\right)\mathrm du\\ &=\int_{e^{-\pi i}\infty}^{-\varepsilon-i\delta}+\int_{-\varepsilon-{ik\over k+k_2}}^{-\varepsilon-{ik\over...
\begin{aligned} &\left|p(n)-\frac1i\int_{y-is}^{y+is}F(iu)e^{2\pi nu}\mathrm du\right| \\ &\ll\exp\left[2\pi\left({1\over24}\cdot{\alpha\over\alpha^2+\beta^2}+\alpha\right)\sqrt n\right] \\ &=\exp\left[2\pi\left(\alpha+{1\over24\alpha}\right)\sqrt n\ri...
On Artin Cokernel of the Quaternion Group Q_{2m} when m=2^h \\cdot p_{1}^{r_1} \\cdot p_{2}^{r_2} \\cdots p_{n}^{r_n} such that p_i are Primes, g.c.d.(p_i, p_j)=1 and p_i eq 2 for all i = 1, 2, ..., n, h and r_i any Positive ...