(1-1*2+1/3-1/4+1/5-.+1/1993-1/1994)*[1/(1+1995)+1/(2+1996)+.+1/(977+2991)]要写算式!
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![(1-1*2+1/3-1/4+1/5-.+1/1993-1/1994)*[1/(1+1995)+1/(2+1996)+.+1/(977+2991)]要写算式!](/uploads/image/z/3960184-40-4.jpg?t=%281-1%2A2%2B1%2F3-1%2F4%2B1%2F5-.%2B1%2F1993-1%2F1994%29%2A%5B1%2F%281%2B1995%29%2B1%2F%282%2B1996%29%2B.%2B1%2F%28977%2B2991%29%5D%E8%A6%81%E5%86%99%E7%AE%97%E5%BC%8F%21)
(1-1*2+1/3-1/4+1/5-.+1/1993-1/1994)*[1/(1+1995)+1/(2+1996)+.+1/(977+2991)]要写算式!
(1-1*2+1/3-1/4+1/5-.+1/1993-1/1994)*[1/(1+1995)+1/(2+1996)+.+1/(977+2991)]
要写算式!
(1-1*2+1/3-1/4+1/5-.+1/1993-1/1994)*[1/(1+1995)+1/(2+1996)+.+1/(977+2991)]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-......+1/1993-1/1994)*[1/(1+1995)+1/(2+1996)+......+1/(977+2991)]
=(1-1*2+1/3-1/4+1/5-......+1/1993-1/1994)*[1/2*(1/998+...1/1994)]
====令
x1=1/1+1/3+...+1/1993
x...
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(1-1*2+1/3-1/4+1/5-......+1/1993-1/1994)*[1/(1+1995)+1/(2+1996)+......+1/(977+2991)]
=(1-1*2+1/3-1/4+1/5-......+1/1993-1/1994)*[1/2*(1/998+...1/1994)]
====令
x1=1/1+1/3+...+1/1993
x2=1/2+1/4+...+1/1994
x3=1/998+1/999+...1/1994
x1+x2 = x2*2+x3
x1-x2 = x3
所以
=====
原式
=1/2*(1-1/2+1/3-1/4+1/5-......+1/1993-1/1994)^2
=====
又
1+1/2+1/3+...+1/(n+1) = ln(n+1) ===>这个公式是极限公式,用错了,只到这里了,牛人继续补上.
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