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）anS（2n

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