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$$ \frac { m ^ { 9 } } { 1 2 n } $$ ...
$$ 1 \cdot 2 ^ { 2 } + 2 \cdot 3 ^ { 2 } + \cdots n ( n + 1 ) ^ { 2 } = \frac { 1 } { 1 2 } n ( n + 1 ) ( a n ^ { 2 } + b n + 1 0 ) $$$ a = \_ , b = \_ . $$ 相关知识点: 整式加减 规律探究与定义新运算 其他规律...
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。 因此Z_{1}~
甚至\[f\left( \vec{a}+\Delta \vec{x} \right)=\sum\limits_{n=0}^{\infty }{\frac{1}{n!}{{\left( \Delta \vec{x}\cdot \frac{\partial }{\partial \vec{a}} \right)}^{n}}f\left( {\vec{a}} \right)}.\] 多元函数的幂级数展开 下面都是关于 \[f\left( {{x}_{1}},...
r_3({{\mathbb {z}}}_p^n) \le (p-\delta _p)^n, [24]. \end{aligned}$$ indeed the argument yields the bound $$\begin{aligned} r_3({{\mathbb {z}}}_p^n) \le (j(p)p)^n, [6] \end{aligned}$$ where $$\begin{aligned} j(p)=\frac{1}{p}\min \limits _{0<t<1...
利用公式:{{1}^{2}}+{{2}^{2}}+{{3}^{2}}+\cdot \cdot \cdot +{{n}^{2}}=\frac{n(n+1)(2n+1)}{6},计算:{{2}^{2}}+{{4}^{2}}+{{6}^{2}}+\cdot \cdot \cdot +{{60}^{2}}= . 相关知识点: 试题来源: 解析 37820 提公因数:{{2}^{2}}+{{4}^...
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 ...
\begin{aligned} T(n) &=\int_0^\infty x^{1/2}e^{-{a^2\over x}-c^2x}\mathrm dx=2\int_0^\infty p^2e^{-{a^2\over p^2}-c^2p^2}\mathrm dp \\ &=-{\mathrm d\over\mathrm dc^2}\int_0^\infty e^{-{a^2\over p^2}-c^2p^2}\mathrm dp:=-\frac1c{\mathrm dJ\...
\begin{aligned} Q_k(n) &=k^32^{-\frac32}\pi^{-\frac32}\lambda_n^{-3}\cdot{B^2\lambda_n^2\over2\pi^\frac12k^2}{\mathrm d\over\mathrm d(B\lambda_n/k)}\left(\cosh(B\lambda_n/k)\over B\lambda_n/k\right) \\ &=k^32^{-\frac32}\pi^{-2}{\mathrm d\over2\lamb...
Answer and Explanation: Given: Sequence: {eq}\frac{1}{2 \cdot 6},\frac{1}{4 \cdot 8}, \frac{1}{6 \cdot 10},\frac{1}{8 \cdot 12}, \dots {/eq} Solution: From the given...Become a member and unlock all ...