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#3

105.5.26補充

[ 本帖最後由 ferng 於 2016-5-28 04:51 PM 編輯 ]

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0<a<1

##### 引用:

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\begin{align} & \left( 1 \right)\ {{x}_{3}}=4 \\ & \left( 2 \right)\ {{x}_{n}}=\frac{n\left( n+1 \right)}{3} \\ & \left( 3 \right)\ \frac{1}{9} \\ \end{align}

\begin{align} & \overline{{{A}_{n-1}}{{M}_{n}}}=\frac{{{x}_{n}}-{{x}_{n-1}}}{2},\overline{{{B}_{n}}{{M}_{n}}}=\frac{\sqrt{3}\left( {{x}_{n}}-{{x}_{n-1}} \right)}{2} \\ & \frac{\sqrt{3}\left( {{x}_{n}}-{{x}_{n-1}} \right)}{2}=\sqrt{\frac{{{x}_{n}}+{{x}_{n-1}}}{2}} \\ & 3{{\left( {{x}_{n}}-{{x}_{n-1}} \right)}^{2}}=2\left( {{x}_{n}}+{{x}_{n-1}} \right)\ \cdots \cdots \left( 1 \right) \\ & 3{{\left( {{x}_{n+1}}-{{x}_{n}} \right)}^{2}}=2\left( {{x}_{n+1}}+{{x}_{n}} \right)\cdots \cdots \left( 2 \right) \\ & \left( 2 \right)-\left( 1 \right) \\ & {{x}_{n+1}}-2{{x}_{n}}+{{x}_{n-1}}=\frac{2}{3} \\ & {{x}_{n+1}}-{{x}_{n}}={{x}_{n}}-{{x}_{n-1}}+\frac{2}{3} \\ & {{x}_{n}}-{{x}_{n-1}}=\frac{2}{3}n \\ & {{x}_{n}}=\frac{n\left( n+1 \right)}{3} \\ \end{align}

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[x]    [cos(-90度)  , -sin(-90度) ]     [x-4]    [4]
[y] = [sin(-90度 )  , cos(-90度) ]  *  [y-0] + [0]  (順旋90度) => x=y+4 , y=-x+4

(8! - 7! * 2) *(1/2) * (1/3)

CD之間一定要有位置，所以有H(3，3)=10種

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