Question

在数列{an}中,已知a1=2,a(n+1)=2an/(an+1),求数列{an}通项公式

详细过程！！！！！！谢谢

Answer 1

a1=2
a2=2a1/(a1+1)=4/3
a3=2a2/(a2+1)=8/3/7/3=8/7
a4=2a3/(a3+1)=16/7/15/7=16/15
...
推测an=2^n/(2^n-1)
数学归纳法证明，简单写下：
假设an成立 即：an=2^n/(2^n-1)
a(n+1)=2an/(an+1)＝[2^(n+1)/(2^n-1)]/[2^n/(2^n-1)+1]
=2^(n+1)/[2^n+2^n-1]
=2^(n+1)/[2^(n+1)-1]
即对于n＋1也成立
所以an=2^n/(2^n-1)

Answer 2

解:其实最简单的方法就是求出前几个数后找规律
a1=2=2/1 a2=4/3 a3=8/7 a4=16/15
所以an=2^n/(2^n - 1)

Answer 3

所以, 2/a(n+1) = (an+1)/an = 1+1/an
设 Bn = 1/an;hence,B1=1/2
则 2B(n+1) = 1+Bn
Bn = (1/2)^n*B1 + (1/2)^n + (1/2)^(n-1) + .....
= (1/2)^(n+1) + (1-0.5^n)/(1-0.5)-1
an = 1/Bn