设 首项为a,公差为d.
则第一项为 a、第二项为 a+d,最后一项为 a+2nd,最后倒数第二项为 a+(2n-1)d.
奇数有 n+1项,偶数有 n项.
奇数项构成 a 为首项、末项为a+2nd、公差为2d 的等差数列
偶数项构成 a+d 为首项、末项为 a+(2n-1)d、2d 为公差的等差数列
S奇 = (a+a+2nd)*(n+1)/2 = (a+nd)(n+1)
S偶 = [a+d+a+(2n-1)d]*n/2 = (a+nd)*n
S奇/S偶 = (n+1)/n
与要证明的结论不同.
为证明题目有错误,举个例子.
设 1 2 3 4 5.
共5项,则 2n+1=5、n=2
S奇 = 1+3+5 = 9
S偶 = 2+4 = 6
S奇/S偶 = 9/6 =3/2
n/(n-1) = 2/1 = 2 ≠ S奇/S偶
(n+1)/n = (2+1)/2 = 3/2 = S奇/S偶
显然 题目错了.
在等差数列中,有2n+1项.则(S奇)除以(S偶)等于n除以(n-1)怎么证明?
在等差数列中,有2n+1项.则(S奇)除以(S偶)等于n除以(n-1)怎么证明?
数学人气:891 ℃时间:2020-01-29 04:45:29
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