回復 1# chwjh32 的帖子
(1) n為奇數
\(\begin{align}
& {{a}_{n}}={{\left( {{a}_{n-1}}+1 \right)}^{2}}-1 \\
& =\frac{1}{4}\left| {{a}_{n-2}}+1 \right|-1 \\
& =\frac{1}{4}{{\left( {{a}_{n-3}}+1 \right)}^{2}}-1 \\
& =\frac{1}{16}\left| {{a}_{n-4}}+1 \right|-1 \\
& =\cdots \cdots \\
& =\frac{1}{{{2}^{n-1}}}\left| {{a}_{1}}+1 \right|-1 \\
& =\frac{1}{{{2}^{n-1}}}-1 \\
\end{align}\)
(2) n為偶數
\(\begin{align}
& {{a}_{n}}=\frac{1}{2}\sqrt{\left| {{a}_{n-1}}+1 \right|}-1 \\
& =\frac{1}{2}\left( {{a}_{n-2}}+1 \right)-1 \\
& =\frac{1}{4}\sqrt{\left| {{a}_{n-3}}+1 \right|}-1 \\
& =\frac{1}{4}\left( {{a}_{n-4}}+1 \right)-1 \\
& =\frac{1}{8}\sqrt{\left| {{a}_{n-5}}+1 \right|}-1 \\
& =\cdots \cdots \\
& =\frac{1}{{{2}^{\frac{n}{2}}}}\sqrt{\left| {{a}_{1}}+1 \right|}-1 \\
& =\frac{1}{{{2}^{^{\frac{n}{2}}}}}-1 \\
\end{align}\)
