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[数列] 一类二阶非线性递推数列

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青青子衿 posted 2015-6-29 14:28 |Read mode
1.\[{a_{n + 1}}^2 + {a_n}^2 + {a_{n - 1}}^2 - 2{a_{n + 1}}{a_n}{a_{n - 1}} = 1\]
2.\[{a_{n + 1}}^2 + {a_n}^2 + {a_{n - 1}}^2 + 4{a_{n + 1}}{a_n}{a_{n - 1}} = 2{a_n}{a_{n - 1}} + 2{a_n}{a_{n + 1}} + 2{a_{n - 1}}{a_{n + 1}}\]

\[\begin{array}{l}
{a_{n + 1}}^2 + {a_n}^2 + {a_{n - 1}}^2 - 2{a_{n + 1}}{a_n}{a_{n - 1}} = 1\\
{( {\frac{{{b_{n + 1}}}}{2}} )^2} + {( {\frac{{{b_n}}}{2}})^2} + {({\frac{{{b_{n - 1}}}}{2}})^2} - \frac{{{b_{n + 1}}{b_n}{b_{n - 1}}}}{4} = 1\\
{b_{n + 1}}^2 + {b_n}^2 + {b_{n - 1}}^2 - {b_{n + 1}}{b_n}{b_{n - 1}} = 4
\end{array}\]

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