∵{an}是各项均为正数的递增的等比数列
∴an=a1*q^(n-1),q>1,a1>0
∴{1/an}为等比数列,公比为1/q
a1+a2+a3=14/3==> a1(1+q+q^2)=14/3 (1)
1/a1+1/a2+1/a3=21/8 ==>1/a1(1+1/q+1/q^2)=21/8 (2)
(1)/(2):a1^2q^2=16/9==>a1q=4/3带回(1)
得:q=2,a1=2/3
∴Sn=2/3(2^n-1)
∵{an}是各项均为正数的递增的等比数列
∴an=a1*q^(n-1),q>1,a1>0
∴{1/an}为等比数列,公比为1/q
a1+a2+a3=14/3==> a1(1+q+q^2)=14/3 (1)
1/a1+1/a2+1/a3=21/8 ==>1/a1(1+1/q+1/q^2)=21/8 (2)
(1)/(2):a1^2q^2=16/9==>a1q=4/3带回(1)
得:q=2,a1=2/3
∴Sn=2/3(2^n-1)