已知数列（log2(An-1)),(n属于N+)为等差数列且A1=3,A3=9求数列(An)的前n项和Sn.

已知数列{bn}={log2 (an-1)}为等差数列，且a1=3 a3=9→
b1=log2 (3-1)=log2(2)=1,b2=log2 (9-1)=log2(8)=3,
公差d=3-1=2,∴bn=1+(n-1)×2,bn=2n-1→
log2 (an-1)=2n-1→
(1).an =2^(2n-1)
(2).a1=2^(1)=2,a2=2^(3)=8,a3=2^5=32,........
an =2^(2n-1),a(n+1) =2^(2n+1)
∴1/（a2-a1)+1/(a3-a2)+…+1/a(n+1)-an=
1/(8-2)+1/(32-8)+......+1/[2^(2n+1)-2^(2n-1)]
=1/6+1/24+......+1/3×(2^(2n-1)
=1/3×2+1/3×2^3+.....1/3×(2^(2n-1)
<1/2+1/2^3+........+1/2^(2n-1)
=(1/2)[1-(1/2)^2n]/[1-(1/2)^2]
=(1/2)[1-(1/2)^2n]/(3/4)
=(2/3)[1-(1/2)^2n]
<2/3<1

我是说噻。
裂项法 详细我就不说了
Sn=2^(n+1)+n-2