设等差数列的公差为d,则d=log2(a2-1)-log2(a1-1)=1∴log2(an-1)=log22+(n-1)×1=n∴an=2n+1则an+1-an=2n+1-2n=2n∴ 1 a2?a1 + 1 a3?a2 +…+ 1 an+1?an = 1 2 + 1 22 +…+ 1 2n = 1 2 [1?( 1 2 )]n 1? 1 2 =1? 1 2n 故答案为:1? 1 2n