(n+1)an,(n+2)an+1,n成等差，则(n+2)an+1=[n+(n+1)an]/2,bn=(n+1)an-n+2,bn+1=(n+2)an+1-n+1,bn+1/bn=1/2所以是等差，b1=-1,bn=-1*(1/2)^n-1bn=(n+1)an-n+2=-1*(1/2)^n-1,an=[-1*(1/2)^n-1+n-2]/(n+1)
