设等差数列｛an｝的公比q＝1/2,前n项和为Sn,则S4/a4＝过程
问题描述：
设等差数列｛an｝的公比q＝1/2,前n项和为Sn,则S4/a4＝
过程
答
s4/a4
=[a1(1-q^4)/(1-q)]/a1q^3
=[(1-q^4)/(1-q)]/q^3
=[(1-q)(1+q)(1+q^2)]/(1-q)]/q^3
=(1+q)(1+q^2)/q^3
=(1+1/2)*(1+1/2^2)/(1/2)^3
=(3/2*5/4)/(1/8)
=15/8*8
=15