第一题:
\[
\begin{array}{l}
\displaystyle
a_j:=\underbrace{7+7+\cdots+7}_{j}=7j
\\
\displaystyle
A_{2k}:=\frac{a_{2k}}{\frac{a_{2k-1}}{\frac{a_{2k-2}}{\frac{\vdots}{a_1}}}}=\frac{\prod_{j=1}^k a_{2j}}{\prod_{j=1}^k a_{2j-1}}=\frac{\cancel{7^k}(2k)!!}{\cancel{7^k} (2k-1)!!}
=\frac{2^{2k} k!^2}{(2k)!}=\frac{2^{2k}}{{2k\choose k}}
\\
\displaystyle
A_8=\frac{2^8}{{8\choose 4}}=\frac{128}{35}
\end{array}
\]
第二题:设 \(x,y\) 分别是待求的长边和短边之长度(实值)。根据复平面上的向量加法规则,有
\[
\begin{array}{ll}
&\displaystyle
x +2 e^{\pi i/3}+9 e^{2\pi i/3}+5 e^{3\pi i/3}+7 e^{4\pi i/3}+y e^{5\pi i/3}=0
\\
\Rightarrow &\displaystyle
x=\frac{24-y}{2}+i\frac{\sqrt{3}(y-4)}{2};\quad\fbox{\({\rm Im}x={\rm Im}y=0\)}~\Rightarrow~\left\{\begin{array}{l}
x=10
\\
y=4
\end{array}\right.
\end{array}
\]
\[
\begin{array}{l}
\displaystyle
a_j:=\underbrace{7+7+\cdots+7}_{j}=7j
\\
\displaystyle
A_{2k}:=\frac{a_{2k}}{\frac{a_{2k-1}}{\frac{a_{2k-2}}{\frac{\vdots}{a_1}}}}=\frac{\prod_{j=1}^k a_{2j}}{\prod_{j=1}^k a_{2j-1}}=\frac{\cancel{7^k}(2k)!!}{\cancel{7^k} (2k-1)!!}
=\frac{2^{2k} k!^2}{(2k)!}=\frac{2^{2k}}{{2k\choose k}}
\\
\displaystyle
A_8=\frac{2^8}{{8\choose 4}}=\frac{128}{35}
\end{array}
\]
第二题:设 \(x,y\) 分别是待求的长边和短边之长度(实值)。根据复平面上的向量加法规则,有
\[
\begin{array}{ll}
&\displaystyle
x +2 e^{\pi i/3}+9 e^{2\pi i/3}+5 e^{3\pi i/3}+7 e^{4\pi i/3}+y e^{5\pi i/3}=0
\\
\Rightarrow &\displaystyle
x=\frac{24-y}{2}+i\frac{\sqrt{3}(y-4)}{2};\quad\fbox{\({\rm Im}x={\rm Im}y=0\)}~\Rightarrow~\left\{\begin{array}{l}
x=10
\\
y=4
\end{array}\right.
\end{array}
\]