回復 25# s7908155 的帖子
第4題
\(\begin{align}
& {{x}_{0}}=0 \\
& {{x}_{1}}=1 \\
& {{x}_{2}}={{x}_{0}}+\frac{1}{3}\left( {{x}_{1}}-{{x}_{0}} \right)=\frac{1}{3}{{x}_{1}}+\frac{2}{3}{{x}_{0}} \\
& {{x}_{3}}={{x}_{2}}+\frac{2}{3}\left( {{x}_{1}}-{{x}_{2}} \right)=\frac{1}{3}{{x}_{2}}+\frac{2}{3}{{x}_{1}} \\
& {{x}_{4}}={{x}_{2}}+\frac{1}{3}\left( {{x}_{3}}-{{x}_{2}} \right)=\frac{1}{3}{{x}_{3}}+\frac{2}{3}{{x}_{2}} \\
& : \\
& : \\
& {{x}_{n}}=\frac{1}{3}{{x}_{n-1}}+\frac{2}{3}{{x}_{n-2}} \\
\end{align}\)