{an}是等比数列,a7=1,a4,a5+1,a6成等差数列,求an和证明前n项和Sn

2(a5+1)=a4+a6
2a5+2=a4+a6
2a1q^4+2=a1q^3+a1q^5
a7=a1q^6=1
a1=1/q^6
2/q^2+2=1/q^3+1/q
2q+2q^3=1+q^2
2q^3-q^2+2q-1=0
2q^3+2q-(q^2+1)=0
2q(q^2+1)-(q^2+1)=0
(q^2+1)(2q-1)=0
2q-1=0
q=1/2
a1=1/(1/2)^6=64
an=a1q^(n-1)=64*(1/2)^(n-1)(n:N*)
Sn=(64(1-(1/2)^n)/(1-1/2)=128(1-(1/2)^n)(n:N*)
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