数列 各项为正数的数列中前N项和SN大于1,且6SN=[an+1]*[an+2]求an通项公式
问题描述：
数列 各项为正数的数列中前N项和SN大于1,且6SN=[an+1]*[an+2]求an通项公式
答
6Sn=(an+1)(an+2)=an*an+3an+26Sn-1=(an-1+1)(an-1+2)=(an-1)*(an-1)+3(an-1)+2相减得6an=an*an-(an-1)*(an-1)+3[an-(an-1)]即[an-a(n-1)][an+a(n-1)]-3[an+a(n-1)]=0即[an-a(n-1)-3][an+a(n-1)]=0得an-a(n-1)=3或an...