设公比为q.
a1+a2=a1(1+q)=9
a1a2a3=a1^3q^^3=27
a1q=3
a1=3/q
(3/q)(1+q)=9
3q+3=9q
6q=3
q=1/2
a1=3/(1/2)=6
Sn=a1(1-q^n)/(1-q)
=6*[1-(1/2)^n]/(1-1/2)
=12[1-(1/2)^n]
=12(2^n-1)/2^n
设公比为q.
a1+a2=a1(1+q)=9
a1a2a3=a1^3q^^3=27
a1q=3
a1=3/q
(3/q)(1+q)=9
3q+3=9q
6q=3
q=1/2
a1=3/(1/2)=6
Sn=a1(1-q^n)/(1-q)
=6*[1-(1/2)^n]/(1-1/2)
=12[1-(1/2)^n]
=12(2^n-1)/2^n