Question

若数列{a_n}满足前n项之和S_n=2a_n-4(n∈N^*),b_n+1=a_n+2b_n,且b₁=2,求:

(1){b_n}的通项公式;

(2){b_n}的前n项和T_n.

Answer 1

解:(1)当n≥2时,a_n=S_n-S_n-1=2a_n-4-2a_n-1+4,

即得a_n=2a_n-1,

当n=1时,a₁=S₁=2a₁-4=4,∴a_n=2ⁿ⁺¹.                                          

∴b_n+1=2ⁿ⁺¹+2b_n.∴=1.

∴{}是以1为首项,以1为公差的等差数列.

∴=1+(n-1)×1=n∴b_n=n·2ⁿ.                                                   

(2)T_n=1·2+2·2²+…+n·2ⁿ,                                                           ①

2T_n=1·2²+2·2³+…+(n-1)·2ⁿ+n·2ⁿ⁺¹,                                       ②

①-②,得-T_n=2+2²+2³+…+2ⁿ-n·2ⁿ⁺¹=n·2ⁿ⁺¹,

∴T_n=(n-1)·2ⁿ⁺¹+2.