设公比为q
(a2+a3)/(a1+a2)=(a1+a2)q/(a1+a2)=q=6/3=2
a1+a2=3 a1(1+q)=3 a1×(2+1)=3 3a1=3
a1=1
an=a1q^(n-1)=1×2^(n-1)=2^(n-1)
an+a(n+1)=2^(n-1)+2ⁿ=3×2^(n-1)
[a(n+1)+a(n+2)]/[an+a(n+1)]=(3×2ⁿ)/[3×2^(n-1)]=2,为定值.
数列{an+a(n+1)}是以3为首项,2为公比的等比数列.
S8=3×(2^8 -1)/(2-1)=3×(256-1)=765