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数学
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an为等比数列 首项为a1 公比为q 前n项和为Sn 求S1+S2+S3+.Sn
人气:279 ℃ 时间:2020-03-19 12:56:10
解答
当q=1时,S1+S2+S3+……+Sn=(1+2+3+……+n)*a1 =n*(n+1)*a1/2
当q不等于1时,Sn=(a(n+1)-a1)/(q-1),所以S1+S2+S3+……+Sn=[(a2+a3+a4+……+a(n+1))-n*a1]/(q-1)=a2*(q^n-1)/(q-1)^2-n*a1/(q-1)
等比数列
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