回復 1# 瓜農自足 的帖子
\(\begin{align}
& {{a}_{1}}=13 \\
& \left( n+1 \right){{a}_{n}}=2{{a}_{n-1}}+n-1 \\
& {{a}_{n}}=\frac{2}{n+1}{{a}_{n-1}}+\frac{n-1}{n+1}=\frac{2}{n+1}{{a}_{n-1}}-\frac{2}{n+1}+1=\frac{2}{n+1}\left( {{a}_{n-1}}-1 \right)+1 \\
& {{a}_{n}}-1=\frac{2}{n+1}\left( {{a}_{n-1}}-1 \right)=\frac{2}{n+1}\times \frac{2}{n}\left( {{a}_{n-2}}-1 \right)=...=\frac{2}{n+1}\times \frac{2}{n}\times ...\times \frac{2}{3}\left( {{a}_{1}}-1 \right)=\frac{{{2}^{n}}}{\left( n+1 \right)!}\times 12 \\
& {{a}_{n}}=\frac{{{2}^{n}}}{\left( n+1 \right)!}\times 12+1 \\
\end{align}\)
