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    科技2022-08-16  99

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    裴波那契数列 a n = a n − 1 + a n − 2 a_n=a_{n-1}+a_{n-2} an=an1+an2 a 1 = 1 a_1=1 a1=1 a 2 = 2 a_2=2 a2=2

    a n + 1 β n + 1 + α β \frac{a_{n+1}}{\beta^{n+1}}+\frac{\alpha}{\beta} βn+1an+1+βα b n = a n β n b_n=\frac{a_n}{\beta^n} bn=βnan

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