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

(1) a₁=1   (2) a_n=3·2^n-1-2

解析解:(1)由题意a₁=S₁=T₁,T_n=2S_n-n²,
令n=1得a₁=2a₁-1,∴a₁=1.
(2)由T_n=2S_n-n²①
得T_n-1=2S_n-1-(n-1)²(n≥2)②
①-②得S_n=2a_n-2n+1(n≥2),
验证n=1时也成立.
∴S_n=2a_n-2n+1③
则S_n-1=2a_n-1-2(n-1)+1(n≥2)④
③-④得a_n=2a_n-2a_n-1-2,
即a_n+2=2(a_n-1+2),
故数列{a_n+2}是公比为2的等比数列,首项为3,
所以a_n+2=3·2^n-1,从而a_n=3·2^n-1-2.