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等比数列{an}中q=1/2且a1+a3+a5+...+a99=60,求s100
人气:464 ℃ 时间:2020-04-01 11:46:08
解答
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
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