Question

设S_n是数列{a_n}(n∈N^*)的前n项和,a₁=1,且S_n²=3n²a_n+S_n-1²,a_n≠0,n=2,3, 4,….

(1)求a₂,a₃的值;

(2)求数列{a_n}的通项公式.

Answer 1

解:(1)当n=2时,由(a₂+1)²=12a₂+1,得a₂(a₂-10)=0,∵a₂≠0,∴a₂=10.

又(11+a₃)²=27a₃+11²,可得a₃(a₃-5)=0,∵a₃≠0,∴a₃=5.

(2)当n≥2时,由已知得S_n²-S_n-1²=3n²a_n,∵a_n=S_n-S_n-1≠0,(S_n-S_n-1)(S_n+S_n-1)=3n²a_n,a_n(S_n+S_n-1)=3n²a_n,

∴S_n+S_n-1=3n²,①

于是S_n+1+S_n=3(n+1)².②

②-①,得a_n+1+a_n=6n+3.③

于是a_n+2+a_n+1=6n+9,④

由④-③得a_n+2-a_n=6,即{a_2k},{a_2k+1}分别是a₂,a₃为首项,以6为公差的等差数列.

∴a_2k=a₂+6(k-1)=6k+4,a_2k+1=a₃+6(k-1)=6k-1,设n=2k,则a_n=3n+4,设n=2k+1,则a_n=3n-4.

∴a_n=