凑项首先,解方程x=(x+4)/(2x+3)得x1=1,x2=-2
[a(n-1)-1]/[a(n-1)+2]=[(an+4)/(2an+3)-1]/[(an+4)/(2an+3)+2]
=(1-an)/(5an+10)=(-1/5)[(an-1)/(an+2)]
则｛(an-1)/(an+2)},是以(3-1)/(3+2)=2/5为首项,公比为q=-5的等比数列
所以(an-1)/(an+2)=(2/5)*(-5)^(n-1)
故an=[5+4*(-5)^(n-1)]/[5-2*(-5)^(n-1)]
来自网上,正确