Question

設數列{an}前n項和為Sn,數列{Sn}的前n項和為Tn,滿足Tn=2Sn-n2,n∈N*.

(1)求a1的值.

(2)求數列{an}的通項公式.

 

Answer 1

(1) a1=1 (2) an=3·2n-1-2,n∈N*

【解析】(1)當n=1時,T1=2S1-1.

因為T1=S1=a1,所以a1=2a1-1,解得a1=1.

(2)當n≥2時,Sn=Tn-Tn-1

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

=2Sn-2Sn-1-2n+1,所以Sn=2Sn-1+2n-1�、�,

所以Sn+1=2Sn+2n+1�、�,

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,則=2.

所以{an+2}是以3為首項,2為公比的等比數列,

所以an+2=3·2n-1,

所以an=3·2n-1-2,n∈N*.