設(shè)公比為q,則q≠0
a4、a5+1、a6成等差數(shù)列,則
2(a5+1)=a4+a6
2(a7/q^2 +1)=a7/q^3 +a7/q
a7=1代入,整理,得
2q^3-q^2+2q-1=0
q^2(2q-1)+(2q-1)=0
(2q-1)(q^2 +1)=0
q^2 +1恒>0,因此只有2q-1=0
q=1/2
an=a1q^(n-1)=a7q^(n-7)=1×(1/2)^(n-7)=1/2^(n-7)
數(shù)列{an}的通項(xiàng)公式為an=1/2^(n-7)
a1=1/2^(1-7)=64
Sn=a1(1-q^n)/(1-q)
=64×(1-1/2^n)/(1-1/2)
=128×(1- 1/2^n)
=128- 1/2^(n-7)
1/2^(n-7)>0 128- 1/2^(n-7)