已知数列{an}满足a1=1,an+1=2an+2.
问题描述:
已知数列{an}满足a1=1,an+1=2an+2.
(1)设bn=2^n/an,求证:数列{bn}是等差数列.
(2)求数列{an}的通项公式.
a(n+1)
答
an+1=2an+2,an=-1,把an=-1代入bn=2^n/an,得,bn=-2^n
b2-b1=-2^*2-(-2)=-6,所以{bn}是等差数列