来自刘向杰的问题
已知等比数列{an}的首项为a1,公比为q,写出其前n项和sn的计算公式并证明Sn=a1+a2+a3+…+an=a1+a1q+a1q2+a1q3+…+a1qn-1则qSn=a1q+a1q2+a1q3+a1q4+…+a1qn两式相减,得(1-q)Sn=(a1+a1q+a1q2+a1q3+…+a1qn-1)-(a1q+a1q2+a1q3+
已知等比数列{an}的首项为a1,公比为q,写出其前n项和sn的计算公式并证明
Sn=a1+a2+a3+…+an=a1+a1q+a1q2+a1q3+…+a1qn-1则qSn=a1q+a1q2+a1q3+a1q4+…+a1qn两式相减,得(1-q)Sn=(a1+a1q+a1q2+a1q3+…+a1qn-1)-(a1q+a1q2+a1q3+a1q4+…+a1qn)(1-q)Sn=a1+a1q-a1q+a1q2-a1q2+a1q3-a1q3+…+a1qn-1-a1qn(1-q)Sn=a1-a1qnSn=a1(1-qn)/(1-q)
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2020-11-18 01:38