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战巡
发表于 2014-1-3 11:33
回复 1# wenshengli
这个还是挺容易的吧，貌似通项都可以求出来
\[a_{n+3}=\frac{1}{a_n}(a_{n+1}a_{n+2}+7)\]
\[a_{n+3}a_n=a_{n+1}a_{n+2}+7\]
\[a_{n+2}a_{n-1}=a_na_{n+1}+7\]
相减可得
\[a_{n+3}a_n-a_{n+2}a_{n-1}=a_{n+1}a_{n+2}-a_na_{n+1}\]
\[\frac{a_{n+3}}{a_{n+2}}-\frac{a_{n-1}}{a_n}=\frac{a_{n+1}}{a_n}-\frac{a_{n+1}}{a_{n+2}}\]
令$\frac{a_{n+1}}{a_n}=b_n$，有
\[b_{n+2}-\frac{1}{b_{n-1}}=b_n-\frac{1}{b_{n+1}}\]
\[b_{n+2}+\frac{1}{b_{n+1}}=b_n+\frac{1}{b_{n-1}}\]
很容易算出$b_1=1, b_2=2, b_3=\frac{9}{2}, b_4=\frac{25}{9}$
可知
\[b_{2n}+\frac{1}{b_{2n-1}}=3\]
\[b_{2n+1}+\frac{1}{b_{2n}}=5\]
\[\frac{a_{2n+1}}{a_{2n}}+\frac{a_{2n-1}}{a_{2n}}=3, a_{2n+1}=3a_{2n}-a_{2n-1}\]
\[\frac{a_{2n+2}}{a_{2n+1}}+\frac{a_{2n}}{a_{2n+1}}=5, a_{2n+2}=5a_{2n+1}-a_{2n}\]
剩下就是数学归纳了
愿意求通项的话也可以做下去，不过我懒得搞了 |
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kuing
发表于 2014-1-3 13:17
