6.已知各项为正的数列{an}\{a_{n}\}{an}的前nnn项和为SnS_{n}Sn,满足Sn=14(an+1)2{S_n}=\frac{1}{4}{({{a_n}+1})^2}Sn=41(an+1)2,则2Sn+6an+3\frac{2{S_n}+6}{{a_n}+3}an+32Sn+6的最小值为((( )))