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令$a_n=\frac{1}{b_n}+b_n(b_0=2),则a_n^2=b_n^2+2+\frac{1}{b_n^2}$
所以$a_{n+1}=a_n^2-2=b_n^2+\frac{1}{b_n^2}$
即$b_{n+1}+\frac{1}{b_{n+1}}=b_n^2+\frac{1}{b_n^2}$
令$n=0,有b_{1}+\frac{1}{b_{1}}=b_0^2+\frac{1}{b_0^2}$
两边平方，$b_0^4+\frac{1}{b_0^4}=b_{1}^2+\frac{1}{b_{1}^2}=b_{2}+\frac{1}{b_{2}}$
以此类推，有$b_0^{2^n}+\frac{1}{b_0^{2^n}}=b_n+\frac{1}{b_n}$
综上有：$a_n=b_n+\frac{1}{b_n}=b_0^{2^n}+\frac{1}{b_0^{2^n}}=2^{2^n}+\frac{1}{2^{2^n}}$ |
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kuing
posted 2018-3-27 11:30
sowhat
posted 2018-4-11 18:45
