在等比数列{an}中,己知S3=13/9,s6=364/9,求a1及an

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在等比数列{an}中,己知S3=13/9,s6=364/9,求a1及an

S6=a1(1-q^6)/(1-q)
S3=a1(1-q³)/(1-q)
S6/S3=(1-q^6)/(1-q³)=1+q³=28
q=3
所以13/9=a1(1-3³)/(1-3)
a1=1/9
an=3^(3-n)

s6-s3=a4+a5+a6=q^3(a1+a2+a3)=q^3*s3
q^3*s3=39
q^3=27
q=3
a1可求易得an同

an=a1q^(n-1)
S3=a1(1+q+q^2)=13/9 (1)
S6 = a1(1+q+q^2+q^3+q^4+q^5) = 364/9 (2)
(2)/(1)
(1+q+q^2+q^3+q^4+q^5)/(1+q+q^2) = 28
q^5+q^4+q^3-27q^2-27q-27=0
let
f(q) = q^5+q^4+q^3-27q^2-27q-27
f(3) = 0
q=3
q=3 , a1=1/9
an = a1q^(n-1) = 3^(n-3)