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

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