等差数列{an}前n项和为Sn,求证S 2n-1项=(2n-1)an
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等差数列{an}前n项和为Sn,求证S 2n-1项=(2n-1)an
等差数列{an}前n项和为Sn,求证S 2n-1项=(2n-1)an
等差数列{an}前n项和为Sn,求证S 2n-1项=(2n-1)an
由等差数列求和公式,sn=(a1+an)*n/2;
得,s(2n-1)=(a1+a(2n-1))*(2n-1)/2;
由中项he公式,a1+a(2n-1)=2*a{(2n-1+1)/2}=2*an;
代入上式即得,s(2n-1)=2*an*(2n-1)/2=(2n-1)an