递推公式 $a_{t}=b_{t}+c_{t}a_{t-1}$ 转通项公式
# 递推公式 $a_{t}=b_{t}+c_{t}a_{t-1}$ 转通项公式
Tags: #Math
$$\begin{aligned} a_{t}=b_{t} &+c_{t} a_{t-1} \\ &+ c_{t}\left(b_{t-1}+c_{t-1} a_{t-2}\right) \\ &\hspace{4.25em}+c_{t-1}\left(b_{t-2}+c_{t-2} a_{t-3}\right) \\ &\hspace{13em}\vdots \\ &\hspace{12.5em}+c_{2}\left(b_{1}+c_{1} a_{0}\right) \\ &\hspace{17.5em}\uparrow\\ &\hspace{17.8em}0\\ \end{aligned}$$ 右边: $$\begin{aligned} &\textcolor{blue}{b_{t}}+\textcolor{darkorange}{c_{t}}(\textcolor{blue}{b_{t-1}}+\cdots \textcolor{darkorange}{c_{4}}(\textcolor{blue}{b_{3}}+\textcolor{darkorange}{c_{3}}(\textcolor{blue}{b_{2}}+\textcolor{darkorange}{c_{2}} \textcolor{blue}{b_{1}})))\\ =&\textcolor{darkorange}{c_{t}c_{t-1}\cdots c_{4}c_{3}c_{2}}\textcolor{blue}{b_{1}}+\textcolor{darkorange}{c_{t}c_{t-1}\cdots c_{4}c_{3}}\textcolor{blue}{b_{2}}+\textcolor{darkorange}{c_{t}c_{t-1}\cdots c_{4}}\textcolor{blue}{b_{3}}+\cdots+\textcolor{darkorange}{c_{t}}\textcolor{blue}{b_{t-1}}+\textcolor{blue}{b_{t}} \=&\textcolor{blue}{b_{t}}+\textcolor{blue}{\sum_{i=1}^{t-1}}\left(\textcolor{darkorange}{\prod_{j=i+1}^{t}c_{j}}\right)\textcolor{blue}{b_{i}} \=&b_{t}+\sum_{i=1}^{t-1}\left(\prod_{j=i+1}^{t}c_{j}\right)b_{i} \end{aligned}$$ 故 $$a_{t}=b_{t}+\sum_{i=1}^{t-1}\left(\prod_{j=i+1}^{t}c_{j}\right)b_{i}$$