Question

设数列{a_n}的前n项和为S_n,数列{S_n}的前n项和为T_n,满足T_n=2S_n-n²,n∈N^*.

(1)求a₁的值;

(2)求数列{a_n}的通项公式.

Answer

(1) a1=1 (2) an=3·2n-1-2

【解析】

解:(1)由题意a1=S1=T1,Tn=2Sn-n2,

令n=1得a1=2a1-1,∴a1=1.

(2)由Tn=2Sn-n2①

得Tn-1=2Sn-1-(n-1)2(n≥2)②

①-②得Sn=2an-2n+1(n≥2),

验证n=1时也成立.

∴Sn=2an-2n+1③

则Sn-1=2an-1-2(n-1)+1(n≥2)④

③-④得an=2an-2an-1-2,

即an+2=2(an-1+2),

故数列{an+2}是公比为2的等比数列,首项为3,

所以an+2=3·2n-1,从而an=3·2n-1-2.