請教遞迴一題！（費氏數列與分項對消）
滿足\(a_1=a_2=1\)，\(a_{n+2}=a_n+a_{n+1}\)的數列\(\langle\;a_n\rangle\;\)稱為費氏數列，則\(\displaystyle \frac{1}{a_1a_3}+\frac{1}{a_2a_4}+\ldots+\frac{1}{a_na_{n+2}}+=\)[u] [/u]。請教大家，感謝！
回復 1# judy75 的帖子
一般項 \(\displaystyle\frac{1}{a_k a_{k+2}}=\left(\frac{1}{a_k}-\frac{1}{a_{k+2}}\right)\frac{1}{a_{k+2}-a_k}\)\(\displaystyle=\left(\frac{1}{a_k}-\frac{1}{a_{k+2}}\right)\frac{1}{a_{k+1}}\)
\(\displaystyle=\frac{1}{a_k a_{k+1}}-\frac{1}{a_{k+1}a_{k+2}}\)
其中 \(k=1,2,3,\cdots\)
因此，\(\displaystyle\lim_{n\to\infty}\sum_{k=1}^{n}\frac{1}{a_k a_{k+2}}=\lim_{n\to\infty}\sum_{k=1}^{n}\left(\frac{1}{a_k a_{k+1}}-\frac{1}{a_{k+1}a_{k+2}}\right)\)
\(\displaystyle=\lim_{n\to\infty}\left(\frac{1}{a_1 a_2}-\frac{1}{a_{n+1}a_{n+2}}\right)\)
易知 \(\displaystyle\lim_{n\to\infty}a_{n+1}=\lim_{n\to\infty}a_{n+2}=\infty\)［可證 \(a_k\geq n, \forall n\geq5\)］
因此，所求＝\(\displaystyle\frac{1}{a_1 a_2}=1\)
回復 2# weiye 的帖子
了解了！感謝您的答覆！辛苦了！頁:
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