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现有甲、乙两个盒子中都有大小、形状、质地相同的 ${\displaystyle 2}$ 个红球和 ${\displaystyle 1}$ 个黑球. 从两个盒子中各取一个球交换，记为一次操作. 重复进行 ${\displaystyle n(n\in {\mathbb{N}}^{\ast })}$ 次操作后，记甲盒子中黑球个数为 ${\displaystyle X_n}$, 甲盒中恰有 ${\displaystyle 1}$ 个黑球的概率为 ${\displaystyle a_n}$, 恰有 ${\displaystyle 2}$ 个黑球的概率为 ${\displaystyle b_n}$.

1. 求随机变量 ${\displaystyle X_1}$ 的分布列;
2. 求数列 ${\displaystyle \{a_n\}}$ 的通项公式;
3. 求证: ${\displaystyle {\displaystyle \sum _{i=1}^n\frac{6-10a_i}{9a_i{a}_{i+1}}<\frac{9}{5}}}$.

初始状态时 ${\displaystyle x_0=1}$, 因此根据对称性，甲盒中任何时刻，有 ${\displaystyle 0,2}$ 个黑球的概率都相等.

1. 解 ${\displaystyle P(X_1=0)={\displaystyle \frac{1}{3}\times \frac{2}{3}=\frac{2}{9}}}$, 因此 ${\displaystyle P(X_1=2)={\displaystyle \frac{2}{9},P(X_1=1)=1-2\times \frac{2}{9}=\frac{4}{9}}}$,因此 ${\displaystyle X_1}$ 的分布列为

   ${\displaystyle X_1}$	${\displaystyle 0}$	${\displaystyle 1}$	${\displaystyle 2}$
   ${\displaystyle P}$	${\displaystyle \frac{2}{9}}$	${\displaystyle \frac{5}{9}}$	${\displaystyle \frac{2}{9}}$

2. 解 由递推关系和概率基本关系$$\begin{array}{r}\{\begin{array}{rl}& a_n+2b_n=1,\\ & b_n=\frac{1}{3}\times \frac{2}{3}{a}_{n-1}+\frac{1}{3}{b}_{n-1},\end{array}\end{array}$$
   消去 ${\displaystyle b_n}$ 得$$\begin{array}{r}{A}_n=-\frac{1}{9}{a}_{n-1}+\frac{2}{3}\Rightarrow a_n-\frac{3}{5}=-\frac{1}{9}(a_{n-1}-\frac{3}{5}),\end{array}$$
   这说明 ${\displaystyle {\displaystyle \{a_n-\frac{3}{5}\}}}$ 是以 ${\displaystyle a_1-{\displaystyle \frac{3}{5}=-\frac{2}{45}}}$ 为首项，${\displaystyle {\displaystyle -\frac{1}{9}}}$ 为公比的等比数列，从而$$\begin{array}{r}{a}_n-\frac{3}{5}=\frac{2}{5}\cdot {(-\frac{1}{9})}^n\Rightarrow a_n=\frac{3}{5}+\frac{2}{5}\cdot {(-\frac{1}{9})}^n.\end{array}$$

3. 证明 代入 ${\displaystyle a_n}$, 且简记 ${\displaystyle q={\displaystyle -\frac{1}{9}}}$,则有$$\begin{array}{rl}\frac{6-10a_i}{9a_i{a}_{i+1}}=& \frac{-4q^i}{9({\displaystyle \frac{3}{5}}+{\displaystyle \frac{2}{5}}{q}^i)({\displaystyle \frac{3}{5}}+{\displaystyle \frac{2}{5}}{q}^{i+1})}=\frac{-4(-{\displaystyle \frac{9}{10}})(q^{i+1}-q^i)}{9({\displaystyle \frac{3}{5}}+{\displaystyle \frac{2}{5}}{q}^i)({\displaystyle \frac{3}{5}}+{\displaystyle \frac{2}{5}}{q}^{i+1})}\\ =& \frac{({\displaystyle \frac{3}{5}}+{\displaystyle \frac{2}{5}}{q}^{i+1})-({\displaystyle \frac{3}{5}}+{\displaystyle \frac{2}{5}}{q}^i)}{({\displaystyle \frac{3}{5}}+{\displaystyle \frac{2}{5}}{q}^i)({\displaystyle \frac{3}{5}}+{\displaystyle \frac{2}{5}}{q}^{i+1})}=\frac{1}{{\displaystyle \frac{3}{5}}+{\displaystyle \frac{2}{5}}{q}^i}-\frac{1}{{\displaystyle \frac{3}{5}}+{\displaystyle \frac{2}{5}}{q}^{i+1}},\end{array}$$
   从而$$\begin{array}{r}\sum _{i=1}^n\frac{6-10a_i}{9a_i{a}_{i+1}}=\frac{1}{{\displaystyle \frac{3}{5}}+{\displaystyle \frac{2}{5}}q}-\frac{1}{{\displaystyle \frac{3}{5}}+{\displaystyle \frac{2}{5}}{q}^{n+1}}<\frac{1}{{\displaystyle \frac{3}{5}}+{\displaystyle \frac{2}{5}}q}=\frac{9}{5}.\end{array}$$