(1)由a(n+2)=3/2a(n+1)-1/2an 得, 2*a(n+2)=3*a(n+1)-an,2*[a(n+2)-a(n+1)]=a(n+1)-an
即2d(n+1)=dn, d(n+1)/dn=1/2
所以：{dn}是等比数列
(2)d1=a2-a1=1/2,{dn}的前n项和为S1=1-(1/2)^n
S1= d1+d2+……+dn=(a2-a1)+(a3-a2)+……+[a(n+1)-an]=a(n+1)-a1=a(n+1)-1=1-(1/2)^n
a(n+1)=2-(1/2)^n ,an=2-(1/2)^(n-1)