Question

數(shù)列{a_n}的前n項(xiàng)和記為S_n,a₁=1,a_n+1=2S_n+1(n∈N₊).

(1)求{a_n}的通項(xiàng)公式;

(2)等差數(shù)列{b_n}的各項(xiàng)為正,其前n項(xiàng)和為T_n,且T₃=15,又a₁+b₁,a₂+b₂,a₃+b₃成等比數(shù)列,求T_n.

Answer 1

解:(1)由a_n+1=2S_n+1可得a_n=2S_n-1+1(n≥2),

兩式相減得a_n+1-a_n=2a_n,a_n+1=3a_n(n≥2).

又a₂=2S₁+1=3,∴a₂=3a₁.∴=3.故{a_n}是首項(xiàng)為1,公比為3的等比數(shù)列.

∴a_n=3^n-1.

(2)設(shè){b_n}的公差為d,由T₃=15,可得b₁+b₂+b₃=15,∴b₂=5.

故可設(shè)b₁=5-d,b₃=5+d.

又a₁=1,a₂=3,a₃=9,由題意可得(5-d+1)(5+d+9)=(5+3)²,解得d₁=2,d₂=-10.

∵等差數(shù)列{b_n}的各項(xiàng)為正，∴d＞0.∴d=2.∴T_n=3n+×2=n²+2n.