等比数列a1+a2=3,a3+a4=6,求S8
问题描述：
等比数列a1+a2=3,a3+a4=6,求S8
答
设公比为p，则a1+a2=(p+1)a1=3,a3+a4=(p+1)a1*p^2=6
得p=√2,a1=3√2-3
s8=(p^8-1)a1/(p-1)=15*(3√2-3）/（√2-1）=45
答
a3+a4=a1q^2+a2q^2=q^2(a1+a2)
6=3q^2
q^2=2
a5+a6
=q^2(a3+a4)
=2*6
=12
a7+a8
=q^2(a5+a6)
=2*12
=24
s8=a1+a2+a3+a4+a5+a6+a7+a8
=3+6+12+24
=45