lga1+lga2+……+lgan
=lga1+lgqa1+lgq^2a1……+lgq^(n-1)a1
=nlga1+lgq+lgq^2+.+lgq^(n-1)
=nlga1+lgq+2lgq+3lgq+.+(n-1)lgq
=nlga1+(1/2)lgq*n(n-1)
关于等比数列
关于等比数列
已知数列{an} 是一个以q（q>0)为公比、以a1(a1>0)为首项的等比数列,求lga1+lga2+……+lgan
已知数列{an} 是一个以q（q>0)为公比、以a1(a1>0)为首项的等比数列,求lga1+lga2+……+lgan
数学人气：422 ℃时间：2020-05-23 05:44:30
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