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-1)
[a(n+1)-an]+[an-a(n-1)]+...+(a2-a1)=a(n+1)-a1=a(n+1)-1
=(-1/3)^(n-1)+(-1/3)^(n-2)+...+(-1/3)+1
=(-1/3)^[1-(-1/3)^(n-1)]/[1-(-1/3)]+1
=(1/4)*[(-1/3)^(n-1)-1]+1
a(n+1)=(1/4)*[(-1/3)^(n-1)-1]+2=(1/4)*(-1/3)^(n-1)+7/4
an=(1/4)*(-1/3)^(n-2)+7/4