Question

设数列{an}满足a1=0且an+1 - an=(1-an+1)(1-an)

设数列{an}满足a1=0且an+1 - an=(1-an+1)(1-an)0分
(1)证明：数列{1/1-an}是等差数列
(2)求{an}的通项公式

Answer 1

(1)
设1/(1 - An) = Bn
1 - An = 1/Bn
1 - An+1 = 1/Bn+1
An = (Bn - 1)/Bn
An+1 = (Bn+1 - 1)/Bn+1

An+1 - An = (Bn+1 - 1)/Bn+1 - (Bn - 1)/Bn
=[Bn(Bn+1 - 1) - Bn+1(Bn - 1)]/BnBn+1
=(Bn+1 - Bn)/BnBn+1

(1 - An+1)(1 - An)=1/Bn * 1/Bn+1 = 1/BnBn+1

(Bn+1 - Bn)/BnBn+1 = 1/BnBn+1
Bn+1 - Bn = 1 = d
B1 = 1/(1 - A1) = 1/(1 - 0) = 1
Bn是首项为1，公差为1的等差数列

(2)
Bn=(n-1)d+B1=n-1+1=n = 1/(1 - An)
1 - An = 1/n
An=1 - 1/n = (n-1)/n

Answer 2

.....