等差数列前n项和为Sn,求证:S2n-1=(2n-1)an
人气:240 ℃ 时间:2020-05-14 05:59:55
解答
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
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