Question

數(shù)列{a_n}中,a₁=0,對(duì)任意n≥2且n∈N^*時(shí){a_n}的各項(xiàng)為正值,a_n+1(a_n+1-2)=4a_n(a_n+1),b_n=log₂(a_n+2)都成立.

(1)求數(shù)列{a_n}、{b_n}的通項(xiàng)公式;

(2)當(dāng)k＞3且k∈N^*時(shí),證明對(duì)任意n∈N^*都有+++…+＞1成立.

Answer 1

（1）解:由a_n+1(a_n+1-2)=4a_n(a_n+1)，得

(a_n+1+2a_n)(a_n+1-2a_n-2)=0.

數(shù)列{a_n}的各項(xiàng)為正值,a_n+1+2a_n＞0,

∴a_n+1=2a_n+2.∴a_n+1+2=2(a_n+2).                                                

又a₁+2=2≠0,

∴數(shù)列{a_n+2}為等比數(shù)列.                                                   

∴a_n+2=(a₁+2)·2^n-1=2ⁿ,a_n=2ⁿ-2,即為數(shù)列{a_n}的通項(xiàng)公式.

b_n=log₂(2ⁿ-2+2)=n.                                                         

(2)證明:設(shè)S=+++…+=+++…+,          ①

∴2S=(+)+(+)+(+)+…+(+).         

當(dāng)x＞0,y＞0時(shí),x+y≥2,

+≥2,

∴(x+y)(+)≥4.

∴+≥,當(dāng)且僅當(dāng)x=y時(shí)等號(hào)成立.                              

上述①式中,k＞3,n＞0,n+1,n+2,…,nk-1全為正,

∴2S＞++…+.

∴S＞＞=2(1-)＞2(1-)=1,得證.