Question

设数列{a_n}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,n∈N*

【解析】(1)当n=1时,T1=2S1-1.

因为T1=S1=a1,所以a1=2a1-1,解得a1=1.

(2)当n≥2时,Sn=Tn-Tn-1

=2Sn-n2-[2Sn-1-(n-1)2]

=2Sn-2Sn-1-2n+1,所以Sn=2Sn-1+2n-1 ①,

所以Sn+1=2Sn+2n+1 ②,

②-①得an+1=2an+2,

所以an+1+2=2(an+2),

即=2(n≥2),

求得a1+2=3,a2+2=6,则=2.

所以{an+2}是以3为首项,2为公比的等比数列,

所以an+2=3·2n-1,

所以an=3·2n-1-2,n∈N*.