(1-1*2+1/3-1/4+1/5-......+1/1993-1/1994)*[1/(1+1995)+1/(2+1996)+......+1/(977+2991)]

来源:百度知道 编辑:UC知道 时间:2024/05/27 22:07:38
要写算式!!!

利用:1-1/2+1/3-1/4……+1/(2n-1)-1/2n=1/(n+1)+1/(n+2)+……1/2n

所以:
(1-1*2+1/3-1/4+1/5-......+1/1993-1/1994)/[1/(1+1995)+1/(2+1996)+......+1/(997+2991)]
=(1/998+....+1/1994)/[1/(1+1995)+1/(2+1996)+......+1/(997+2991)]
=(1/2)*(1/998+....+1/1994)/[1/998+1/999+......+1/1994]
=1/2

补充:
1
题目有点问题,最后应该是997+2991 否则和前面规律对不上了!
1 2 3.. 997
1995 1996....2991
2
还是有问题,中间为除,不为乘!

3
关于:
1-1/2+1/3-1/4……+1/(2n-1)-1/2n=1/(n+1)+1/(n+2)+……1/2n
可以用数学归纳法证明:

如下:

当n=1时,左侧=1-1/2=1/2,右侧=1/2,结论成立;
假设n=k成立,则1-1/2+1/3-1/4……+1/(2k-1)-1/2k=1/(k+1)+1/(k+2)+……1/2k
当n=k+1时,左侧={1-1/2+1/3-1/4……+1/(2k-1)-1/2k}+1/(2k+1)-1/(2k +2)
右侧=1/(k+2)+……1/2k+1/(2k+1)+1/(2k +2)={1/(k+1)+1/(k+2)+……1/2k}+1/(2k+1)+1/(2k +2)-1/(k+1)=)={1/(k+1)+1/(k+2)+……1/2k}+1/(2k+1)-1/(2k +2)

根据假设,所以当n=k+1时,左侧=右侧,

所以1-1/2+1/3-1/4……+1/(2n-1)-1/2n=1/(n+1)+1/(n+2)+……1/2n

(1-1*2+1/3-1/4+1/5-......+