已知等比数列{an}的首项a1=2,公比q=3,Sn是它的前n项和.求证：Sn+1/Sn

Sn=2(1-3^n)/(1-3)=3^n-1
S(n+1)=3*3^n-1
S(n+1)/Sn=(3*3^n-1)/(3^n-1)=(3*3^n-3+2)/(3^n-1)=3+2/(3^n-1)
(3n+1)/n =3+1/n
证明原式只需证明2/(3^n-1)≤1/n
即证明(3^n-1)≥2n
设y=3^n-2n-1
当n=1时,y=0,
当n从1增大时,3^n不2n增加的快,所以y≥0
即3^n-2n-1≥0,
3^n-1>2n
原式得证