等比数列{an}中q=1/2且a1+a3+a5+...+a99=60,求s100

来源：百度知道 编辑：UC知道 时间：2024/06/20 17:57:27

a3=a1*q^2, ^2表示平方
a5=a1*q^4...
a1+a3+a5+...+a99
=a1(1+q^2+...+q^98)
=a1(1-q^50)/(1-q^2)
=a1(1-q^50)/[1-q][1+q]
=60

S100=a1(1-q^100)/(1-q)
=a1[(1-q^50)(1+q^50)]/(1-q)
=(1+q^50)(1+q) *[a1[(1-q^50)/(1-q)(1+q）]
=(1+q^50)(1+q)*60
=(1+1/2^50)(1+1/2)*60
=90+90/2^50

a1+a3+a5+...+a99=60
两边同乘以q可得a2+a4+a6...+a100=60*1/2=30
s100=60+30=90