还有已知项数为2n+1等差数列,S奇-S偶=?S奇/S偶=?
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还有已知项数为2n+1等差数列,S奇-S偶=?S奇/S偶=?
![还有已知项数为2n+1等差数列,S奇-S偶=?S奇/S偶=?](/uploads/image/z/16601618-2-8.jpg?t=%E8%BF%98%E6%9C%89%E5%B7%B2%E7%9F%A5%E9%A1%B9%E6%95%B0%E4%B8%BA2n%2B1%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%2CS%E5%A5%87-S%E5%81%B6%3D%3FS%E5%A5%87%2FS%E5%81%B6%3D%3F)
奇数项有n+1项,偶数项有n项
奇数项、偶数项分别成等差数列
S奇=(A1+A(2n+1))×(n+1)/2
=(A1+A1+2nd)×(n+1)/2
=(A1+nd)×(n+1)
=(n+1)A(n+1)
S偶=(A2+A(2n))×n/2
=(A1+d+A1+(2n-1)d)×n/2
=(A1+nd)×n
=nA(n+1)
S奇-S偶=(n+1)A(n+1)-nA(n+1)=A(n+1)
S奇/S偶=(n+1)A(n+1)/nA(n+1)=(n+1)/n
奇数项、偶数项分别成等差数列
S奇=(A1+A(2n+1))×(n+1)/2
=(A1+A1+2nd)×(n+1)/2
=(A1+nd)×(n+1)
=(n+1)A(n+1)
S偶=(A2+A(2n))×n/2
=(A1+d+A1+(2n-1)d)×n/2
=(A1+nd)×n
=nA(n+1)
S奇-S偶=(n+1)A(n+1)-nA(n+1)=A(n+1)
S奇/S偶=(n+1)A(n+1)/nA(n+1)=(n+1)/n
等差数列{An},项数为2n,为何 S奇/S偶 = (An+1)/An?
若等差数列{An}项数为2n,则S偶-S奇=nd,S奇/S偶=An/An-1为什么?
等差数列的项数为2NS偶-S奇=ND,S奇/S偶=an/an+1
项数为(2n-1)时 ,求S偶-S奇=?S偶/S奇=?
证明.项数为奇数2n-1的等差数列{an},有 S奇-S偶=an,s奇/S偶=n/n-1.
等差数列,当项数为2n+1,如何推导S奇-S偶=a1+nd
项数为2n-1项,求证S奇/S偶=n/n-1!
证明.项数为奇数2n的等差数列{an},有 S奇-S偶=an,s奇/S偶=n/n-1.
若等差数列的项数为2n,则S2n=n(an+an+1)与S偶-S奇=nd,S奇分之S偶=an分之an+1怎么得到的。
数学证明题:若等差数列的项数为2n-1(n∈N+),则S奇/S偶=n/(n-1).
数列性质证明问题项数为奇数2n-1的等差数列{an}中 有一个性质是S奇-S偶=an (过程)S奇-S偶=(a1-a2)
求证:当等差数列{an}中的项数为2n-1时,S奇-S偶=an (n为下标)