设Sn是数列{an}（n属于N）的前n项和 a1=a 且 SN^2=3n^2an+Sn^2 下脚标n-1 an不等于0 n=2 3 4 ....

如何你的条件是(S_n)^2 = 3n^2 a_n + (S_{n-1})^2，那么解答是这样的：
3n^2 a_n = (S_n)^2 - (S_{n-1})^2 = (S_n - S_{n-1})(S_n + S_{n-1}) = a_n (S_n + S_{n-1})
于是由a_n≠0，3n^2 = (S_n + S_{n-1})①
从而3(n+1)^2 = (S_{n+1} + S_n)②
①②两式相减：6n+3 = S_{n+1} - S_{n-1} = a_{n+1} + a_n③
进而：6(n+1) + 3 = a_{n+2} + a_{n+1}④
再把③④两式相减：6 = a_{n+2} - a_n