已知AN中,a1=2,6Sn=（an+1)(an+2),求数列An的通项公式An和Sn

6Sn=（an+1)(an+2)6Sn-1=（an-1+1)(an-1+2),相减得：6an=（an+an-1）（an-an-1）+3（an-an-1）继续化简：（an+an-1）（an-an-1-3）=0 ④是不是少了条件,an应该是正数吧? ④中：（an+an-1）不可能为0,所以：（a...