等差数列前n项和为Sn,求证:S2n-1=(2n-1)an

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等差数列前n项和为Sn,求证:S2n-1=(2n-1)an等差数列前n项和为Sn,求证:S2n-1=(2n-1)an等差数列前n项和为Sn,求证:S2n-1=(2n-1)an证：设公差为dS(2n-1

等差数列前n项和为Sn,求证:S2n-1=(2n-1)an
等差数列前n项和为Sn,求证:S2n-1=(2n-1)an

等差数列前n项和为Sn,求证:S2n-1=(2n-1)an
证：
设公差为d
S(2n-1)=[a1+a(2n-1)](2n-1)/2
=[an-(n-1)d+an+(n-1)d](2n-1)/2
=(2an)(2n-1)/2
=(2n-1)an
如果学过等差中项,连公差d都不用设了,an是a1与a(2n-1)的等差中项.
证：
S(2n-1)=[a1+a(2n-1)](2n-1)/2
=(2an)(2n-1)/2
=(2n-1)an

用逆推法可得S2n-1=（a1+a2n-1)*(2n-1)/2
=2an/2*(2n-1)=an(2n-1)

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