求解过程
原题
a k = 1 − β ( 1 − α a k − 2 ) , k = 2 n + 1 , n ∈ Z , α , β ∈ [ 0 , 1 ] a_k = 1-\beta(1-\alpha a_{k-2}), k=2n+1,n\in Z, \alpha,\beta\in [0,1] ak=1−β(1−αak−2),k=2n+1,n∈Z,α,β∈[0,1]
转换问题
a n = 1 − β ( 1 − α a n − 1 ) = ( 1 − β ) + α β a n − 1 , n ∈ Z , α , β ∈ [ 0 , 1 ] a_n = 1-\beta(1-\alpha a_{n-1})=(1-\beta)+\alpha\beta a_{n-1},n\in Z, \alpha,\beta\in [0,1] an=1−β(1−αan−1)=(1−β)+αβan−1,n∈Z,α,β∈[0,1]
解法一
递推:
a n = ( 1 − β ) + α β a n − 1 = ( 1 − β ) + α β ( ( 1 − β ) + α β a n − 2 ) = ( 1 − β ) + α β ( 1 − β ) + α 2 β 2 a n − 2 = ( 1 − β ) + α β ( 1 − β ) + α 2 β 2 ( ( 1 − β ) + α β a n − 3 ) = ( 1 − β ) + α β ( 1 − β ) + α 2 β 2 ( 1 − β ) + α 3 β 3 a n − 3 = . . . = ( 1 − β ) ( 1 + α β + α 2 β 2 + . . . + α n − 2 β n − 2 ) + α n − 1 β n − 1 ∗ a 1 \begin{aligned}a_n &=(1-\beta)+\alpha\beta a_{n-1}\\ &=(1-\beta)+\alpha\beta ((1-\beta)+\alpha\beta a_{n-2})\\ &=(1-\beta)+\alpha\beta (1-\beta)+\alpha^2\beta^2 a_{n-2}\\ &=(1-\beta)+\alpha\beta (1-\beta)+\alpha^2\beta^2 ((1-\beta)+\alpha\beta a_{n-3})\\ &=(1-\beta)+\alpha\beta (1-\beta)+\alpha^2\beta^2 (1-\beta)+\alpha^3\beta^3 a_{n-3}\\ &=...\\ &=(1-\beta)(1+\alpha\beta+\alpha^2\beta^2+...+\alpha^{n-2}\beta^{n-2})+\alpha^{n-1}\beta^{n-1}*a_1\\ \end{aligned} an=(1−β)+αβan−1=(1−β)+αβ((1−β)+αβan−2)=(1−β)+αβ(1−β)+α2β2an−2=(1−β)+αβ(1−β)+α2β2((1−β)+αβan−3)=(1−β)+αβ(1−β)+α2β2(1−β)+α3β3an−3=...=(1−β)(1+αβ+α2β2+...+αn−2βn−2)+αn−1βn−1∗a1
根据等比数列求和公式,
1 + α β + α 2 β 2 + . . . + α n − 2 β n − 2 ( 共有 n − 1 项 ) = 1 − α n − 1 β n − 1 1 − α β 1+\alpha\beta+\alpha^2\beta^2+...+\alpha^{n-2}\beta^{n-2}(共有n-1项)=\frac{1-\alpha^{n-1}\beta^{n-1}}{1-\alpha\beta} 1+αβ+α2β2+...+αn−2βn−2(共有n−1项)=1−αβ1−αn−1βn−1
所以
a n = ( 1 − β ) ( 1 + α β + α 2 β 2 + . . . + α n − 2 β n − 2 ) + α n − 1 β n − 1 ∗ a 1 = ( 1 − β ) 1 − α n − 1 β n − 1 1 − α β + α n − 1 β n − 1 \begin{aligned}a_n &=(1-\beta)(1+\alpha\beta+\alpha^2\beta^2+...+\alpha^{n-2}\beta^{n-2})+\alpha^{n-1}\beta^{n-1}*a_1\\ &=(1-\beta)\frac{1-\alpha^{n-1}\beta^{n-1}}{1-\alpha\beta}+\alpha^{n-1}\beta^{n-1}\\ \end{aligned} an=(1−β)(1+αβ+α2β2+...+αn−2βn−2)+αn−1βn−1∗a1=(1−β)1−αβ1−αn−1βn−1+αn−1βn−1
由于 α , β < 1 \alpha,\beta<1 α,β<1,则
lim n → ∞ α n − 1 β n − 1 = 0 lim n → ∞ ( 1 − α n − 1 β n − 1 ) = 1 \lim_{n\to\infty}{\alpha^{n-1}\beta^{n-1}}=0\\ \lim_{n\to\infty}{(1-\alpha^{n-1}\beta^{n-1})}=1 n→∞limαn−1βn−1=0n→∞lim(1−αn−1βn−1)=1
因此
lim n → ∞ a n = 1 − β 1 − α β \lim_{n\to\infty}a_n=\frac{1-\beta}{1-\alpha\beta} n→∞liman=1−αβ1−β
解法二
易证该数列的极限存在。设极限为x,则x满足:
x = ( 1 − β ) + α β x x=(1-\beta)+\alpha\beta x x=(1−β)+αβx
解得
x = 1 − β 1 − α β x=\frac{1-\beta}{1-\alpha\beta} x=1−αβ1−β