Question

a1=3,2a(n+1)=an+1,求通项公式an

Answer 1

2(a(n+1)+d)=an+d

2a(n+1)=an-d

故d=-1

2（a(n+1)-1)=an-1

a(n+1)-1=1/2(an-1)

于是数列{an-1}是首项a1-1=3-1=2,公比q=1/2的等比数列

故an-1=2*(1/2)^(n-1)=(1/2)^(n-2)

an=(1/2)^(n-2)+1