a(n+1)=-2an+3a(n+1)+k=-2an+3+k=-2(an-3/2-k/2)则令k=-3/2-k/2k=-1则两边同时加-1a(n+1)-1=-2(an-1)[a(n+1)-3]/(an-3)=-2所以{an-3}是等比数列,q=-2a1-3=-1所以an-1=-1*(-2)^(n-1)an=-(-2)^(n-1)+1...