若{an}是等比数列,公比为q,前项和为Sn,求证Sn=a1(1-q^n)/(1-q)
证明:Sn=a1+a2+a3+...+an=a1+a1q+a1q^2+a1q^3+...+a1q^(n-1). (1) 则
qSn=a1q+a1q^2+a1q^3+...+a1q^(n-1)+a1q^n (2)
(1)-(2)得(1-q)Sn=a1-a1q^n=a1(1-q^n).所以Sn=a1(1-q^n)/(1-q)
S=a1+a2+a3+····+an
=a1+a1·q+a1·q2+····+a1·q(n-1)
=a1(1+q+q2+···+q(n-1))
S*q=a1(q+q2+q3+·····+qn)
S(1-q)=a1(1-qn)
S=a1(1-qn)/(1-q)
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