等差数列an前n项和为Sn,求证S(2n-1)=(2n-1)an

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等差数列an前n项和为Sn,求证S(2n-1)=(2n-1)an
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等差数列an前n项和为Sn,求证S(2n-1)=(2n-1)an
等差数列an前n项和为Sn,求证S(2n-1)=(2n-1)an

等差数列an前n项和为Sn,求证S(2n-1)=(2n-1)an
S(2n-1)=(a1+a2n-1)(2n-1)/2=(2n-1)[a1+a1+(2n-2)d]/2=(2n-1)(a1+(n-1)d)=(2n-1)an