Question

已知数列{a_n}的前n项和为S_n,且S_n=2n²+n,n∈N^*,数列{b_n}满足a_n=4log₂b_n+3,n∈N^*.

(1)求a_n,b_n;

(2)求数列{a_n·b_n}的前n项和T_n.

Answer

(1) an=4n-1,n∈N* bn=2n-1,n∈N* (2) Tn=(4n-5)2n+5,n∈N*

【解析】

解:(1)由Sn=2n2+n,得

当n=1时,a1=S1=3;

当n≥2时,an=Sn-Sn-1=4n-1

所以an=4n-1,n∈N*

由4n-1=an=4 log2bn+3,得bn=2n-1,n∈N*.

(2)由(1)知anbn=(4n-1)·2n-1,n∈N*,

所以Tn=3+7×2+11×22+…+(4n-1)·2n-1,

2Tn=3×2+7×22+…+(4n-5)·2n-1+(4n-1)·2n,

所以2Tn-Tn=(4n-1)2n-[3+4(2+22+…+2n-1)]=(4n-5)2n+5

故Tn=(4n-5)2n+5,n∈N*.