数列{{an}中,a1=1,a2=2,3a(n+2)=2a(n+1)+an,求数列{an}的通项公式

3a(n+2)=2a(n+1)+an则3a(n+2)-3a(n+1)=-a(n+1)+an则3【a(n+2)-a(n+1)】=-【a(n+1)-an】【a(n+2)-a(n+1)】/【a(n+1)-an】=-1/3{a(n+1)-an}为q=-1/3的等比数列a(n+1)-an=(a2-a1)*q^(n-1)=(2-1)*(-1/3)^(n-1)=(-1/3)^(n...