2a(n+1)-an=n-2/n(n+1)(n+2)2a(n+1)-2/(n+1)(n+2)=an-1/n(n+1)[a(n+1)-1/(n+1)(n+2)]/[an-1/n(n+1)]=1/2bn=an-1/n(n+1),{bn}是等比数列,公比为1/2,首项为1/2bn=an-1/n(n+1)=(1/2)^nan=(1/2)^n+1/n(n+1)