${{M}_{n}}$ 为满足以下条件的 $n\times n$ 矩阵：对 $1\leqslant i\leqslant n\text{,}{{m}_{i\text{,}i}}\text{=10;}$ 对 $1\leqslant i\leqslant n-1\text{,}{{m}_{i+1\text{,}i}}\text{=}{{m}_{i\text{,i}+1}}\text{=}3\text{;}$ 其余元素为 $0$ 。令 ${{D}_{n}}$ 为 ${{M}_{n}}$ 行列式的值。记 $\displaystyle \sum\limits_{n\text{=}1}^{\infty }{\frac{1}{8{{D}_{n}}+1}\text{=}}\frac{p}{q}$，其中 $p\text{,}q$ 为互质正整数。求 $p+q$ 。
【难度】
【出处】
2011年第29届美国数学邀请赛Ⅱ（AIMEⅡ）
【标注】
【答案】
073
【解析】
$\begin{align}
& {{D}_{1}}=\left| 10\right|=10,{{D}_{2}}=\left| \begin{matrix}
10 & 3 \\
3 & 10 \\
\end{matrix}\right|=\left( 10 \right)\left( 10 \right)-\left( 3 \right)\left( 3 \right)=91\\
& {{D}_{3}}=\left| \begin{matrix}
10 & 3 & 0 \\
3 & 10 & 3 \\
0 & 3 & 10 \\
\end{matrix}\right|=10\left| \begin{matrix}10 & 3 \\
3& 10 \\
\end{matrix}\right|-3\left| \begin{matrix}3 & 3 \\
0 & 10 \\
\end{matrix}\right|+0\left| \begin{matrix}3 & 10 \\
0 & 3 \\
\end{matrix}\right|=10{{D}_{2}}-9{{D}_{1}}=820 \\
&{{D}_{n}}\text{=}10{{D}_{n\text{-}1}}-9{{D}_{n\text{-}2}}\left( n\text{}2\right) \\
\end{align}$，数列满足递推规律，还可写作 ${{D}_{n}}\text{=}10\left({{D}_{n\text{-}1}}\text{-}{{D}_{n\text{-}2}} \right)+{{D}_{n\text{-}2}}$，于是 ${{D}_{0}}\text{=}1{{D}_{4}}=7381$ 。我们观察相邻两项之间的差，$\displaystyle \begin{align}&{{D}_{0}}=1={{9}^{0}},{{D}_{1}}-{{D}_{0}}=10-1={{9}^{1}},{{D}_{2}}-{{D}_{1}}=91-10=81={{9}^{2}},\\
&{{D}_{3}}-{{D}_{2}}=820-91=81=729={{9}^{3}},{{D}_{4}}-{{D}_{3}}=7381-820=6561={{9}^{4}},\\
& {{D}_{n}}\text{=}{{D}_{0}}+{{9}^{0}}+{{9}^{1}}+\cdots+{{9}^{n}}=\sum\limits_{i\text{=}0}^{n}{{{9}^{i}}}\text{=}\frac{\left( 1\right)\left( {{9}^{n+1}}-1 \right)}{9-1}\text{=}\frac{{{9}^{n+1}}-1}{8} \\
\end{align}$ 所求和式为 $\displaystyle \sum\limits_{n\text{=}1}^{\infty}{\frac{1}{8\frac{{{9}^{n+1}}}{8}+1}\text{=}\sum\limits_{n\text{=}1}^{\infty}{\frac{1}{{{9}^{n+1}}-1+1}\text{=}\sum\limits_{n\text{=}1}^{\infty}{\frac{1}{{{9}^{n+1}}}\text{=}\frac{1}{72}}}}$ 所以所求答案为 $073$
& {{D}_{1}}=\left| 10\right|=10,{{D}_{2}}=\left| \begin{matrix}
10 & 3 \\
3 & 10 \\
\end{matrix}\right|=\left( 10 \right)\left( 10 \right)-\left( 3 \right)\left( 3 \right)=91\\
& {{D}_{3}}=\left| \begin{matrix}
10 & 3 & 0 \\
3 & 10 & 3 \\
0 & 3 & 10 \\
\end{matrix}\right|=10\left| \begin{matrix}10 & 3 \\
3& 10 \\
\end{matrix}\right|-3\left| \begin{matrix}3 & 3 \\
0 & 10 \\
\end{matrix}\right|+0\left| \begin{matrix}3 & 10 \\
0 & 3 \\
\end{matrix}\right|=10{{D}_{2}}-9{{D}_{1}}=820 \\
&{{D}_{n}}\text{=}10{{D}_{n\text{-}1}}-9{{D}_{n\text{-}2}}\left( n\text{}2\right) \\
\end{align}$，数列满足递推规律，还可写作 ${{D}_{n}}\text{=}10\left({{D}_{n\text{-}1}}\text{-}{{D}_{n\text{-}2}} \right)+{{D}_{n\text{-}2}}$，于是 ${{D}_{0}}\text{=}1{{D}_{4}}=7381$ 。我们观察相邻两项之间的差，$\displaystyle \begin{align}&{{D}_{0}}=1={{9}^{0}},{{D}_{1}}-{{D}_{0}}=10-1={{9}^{1}},{{D}_{2}}-{{D}_{1}}=91-10=81={{9}^{2}},\\
&{{D}_{3}}-{{D}_{2}}=820-91=81=729={{9}^{3}},{{D}_{4}}-{{D}_{3}}=7381-820=6561={{9}^{4}},\\
& {{D}_{n}}\text{=}{{D}_{0}}+{{9}^{0}}+{{9}^{1}}+\cdots+{{9}^{n}}=\sum\limits_{i\text{=}0}^{n}{{{9}^{i}}}\text{=}\frac{\left( 1\right)\left( {{9}^{n+1}}-1 \right)}{9-1}\text{=}\frac{{{9}^{n+1}}-1}{8} \\
\end{align}$ 所求和式为 $\displaystyle \sum\limits_{n\text{=}1}^{\infty}{\frac{1}{8\frac{{{9}^{n+1}}}{8}+1}\text{=}\sum\limits_{n\text{=}1}^{\infty}{\frac{1}{{{9}^{n+1}}-1+1}\text{=}\sum\limits_{n\text{=}1}^{\infty}{\frac{1}{{{9}^{n+1}}}\text{=}\frac{1}{72}}}}$ 所以所求答案为 $073$
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