a1+a1=1
a1=1/2
an+sn=n
an-1+sn-1=n-1
an-an-1+sn-sn-1=n-(n-1)
2an=a(n-1)+1
2[an-1]=a(n-1)-1
(an-1)/[a(n-1)-1]=1/2
{an-1}等比數(shù)列公比1/2；首項(xiàng)1/2-1=-1/2
an-1=-1/2*(1/2)^(n-1)=-(1/2)^n
bn=an-(an-1) =-(1/2)^n+(1/2)^(n-1)=(1/2)^(n-2)