Question

已知數(shù)列{a_n}的前n項和為S_n,且S_n=2n²+n,n∈N^*,數(shù)列{b_n}滿足a_n=4log₂b_n+3,n∈N^*.
(1)求a_n,b_n;
(2)求數(shù)列{a_n·b_n}的前n項和T_n.

Answer 1

(1) a_n=4n-1,n∈N^*  b_n=2^n-1,n∈N^*  (2) T_n=(4n-5)2ⁿ+5,n∈N^*

解析解:(1)由S_n=2n²+n,得
當(dāng)n=1時,a₁=S₁=3;
當(dāng)n≥2時,a_n=S_n-S_n-1=4n-1
所以a_n=4n-1,n∈N^*
由4n-1=a_n="4" log₂b_n+3,得b_n=2^n-1,n∈N^*.
(2)由(1)知a_nb_n=(4n-1)·2^n-1,n∈N^*,
所以T_n=3+7×2+11×2²+…+(4n-1)·2^n-1,
2T_n=3×2+7×2²+…+(4n-5)·2^n-1+(4n-1)·2ⁿ,
所以2T_n-T_n=(4n-1)2ⁿ-[3+4(2+2²+…+2^n-1)]=(4n-5)2ⁿ+5
故T_n=(4n-5)2ⁿ+5,n∈N^*.