sn为数列{an}的前n项和.已知an>0,an²+2an=4Sn+3

n≥2时， an²+2an=4Sn+3 a(n-1)²+2a(n-1)=4S(n-1)+3 an²+2an-a(n-1)²-2a(n-1)=4[Sn-S(n-1)]=4an an²-a(n-1)²-2an-2a(n-1)=0 [an+a(n-1)][an-a(n-1)]-2[an+a(n-1)]=0 [an+a(n-1)][an-a(n-1)-2]=0 an>0，an+a(n-1)恒>0，因此只有an-a(n-1)-2=0 an-a(