数列an满足a1=2,3a(n+1)+an-7=0,求数列an的通项公式

a1=2,3a(n+1)+an-7=0 (1)设a(n+1)+k=-1/3(an+k)a(n+1)+k=-1/3an-k/33a(n+1)=-an-k-3k-4k=7k=-7/4a1-7/4=2-7/4=1/4{an-7/4}是以1/4首项,-1/3的等比数列an-7/4=1/4*(-1/3)^(n-1)an=1/4*(-1/3)^(n-1)+7/4