Question

設(shè)數(shù)列{a_n}前n項和為S_n,數(shù)列{S_n}的前n項和為T_n,滿足T_n=2S_n-n²,n∈N^*.
(1)求a₁的值.
(2)求數(shù)列{a_n}的通項公式.

Answer 1

(1) a₁=1    (2) a_n=3·2^n-1-2,n∈N^*

(1)當(dāng)n=1時,T₁=2S₁-1.
因?yàn)門₁=S₁=a₁,所以a₁=2a₁-1,解得a₁=1.
(2)當(dāng)n≥2時,S_n=T_n-T_n-1
=2S_n-n²-[2S_n-1-(n-1)²]
=2S_n-2S_n-1-2n+1,所以S_n=2S_n-1+2n-1�、�,
所以S_n+1=2S_n+2n+1�、�,
②-①得a_n+1=2a_n+2,
所以a_n+1+2=2(a_n+2),
即=2(n≥2),
求得a₁+2=3,a₂+2=6,則=2.
所以{a_n+2}是以3為首項,2為公比的等比數(shù)列,
所以a_n+2=3·2^n-1,
所以a_n=3·2^n-1-2,n∈N^*.