香港票房历史前50:1+1/(1+2)+1/(1+2+3)+1/(1+2+3+4)+...1/(1+2+3+...99+100)

来源：百度文库编辑：查人人中国名人网时间：2024/04/30 09:13:24

要详解，谢谢

注意观察，第 N 个加式可以表述成：

1/(1 + 2 + 3 + ... + n)
= 1/[n(n + 1)/2]
= 2/[n(n + 1)]
= 2[1/n - 1/(n + 1)]

那么有：

1/1 + 1/(1 + 2) + 1/(1 + 2 + 3) + .... + 1/(1 + 2 + 3 + ... + 100)
= 1 + 2/(2*3) + 2/(3*4) + .... + 2/(100*101)
= 1 + 2*(1/2 - 1/3) + 2*(1/3 - 1/4) + ... + 2*(1/100 - 1/101)
= 1 + 2*(1/2 - 1/101)
= 2 - 2/101
= 200/101

1+2+...+n=n(n+1)/2
1/(1+2+...+n)=2/[n(n+1)]=2[1/n-1/(n+1)]
原式=2(1-1/2+1/2-1/3+...+1/100-1/101)
=2(1-1/101)
=200/101

1+2+...+n = n*(n+1)/2
则 1/(1+2+...n) = 2/(n*(n+1)) = 2*(1/n-1/(n+1))
故原式=1+2*(1/2-1/3+1/3-1/4+...-1/100+1/101)
=1+2*(1/2-1/101) = 1+99/101 = 200/101

1/(1 + 2 + 3 + ... + n)
= 1/[n(n + 1)/2]
= 2/[n(n + 1)]
= 2[1/n - 1/(n + 1)]

1/1 + 1/(1 + 2) + 1/(1 + 2 + 3) + .... + 1/(1 + 2 + 3 + ... + 100)
= 1 + 2/(2*3) + 2/(3*4) + .... + 2/(100*101)
= 1 + 2*(1/2 - 1/3) + 2*(1/3 - ¼) + ... + 2*(1/100 - 1/101)
= 1 + 2*(1/2 - 1/101)
= 2 - 2/101
= 200/101

1+2+...+n=n(n+1)/2
1/(1+2+...+n)=2/[n(n+1)]=2[1/n-1/(n+1)]
原式=2(1-1/2+1/2-1/3+...+1/100-1/101)
=2(1-1/101)
=200/101

随意挑修。。。

(1/2005-1)(1/2004-1)........(1/3-1)(1/2-1) 1+1/1+2/=?? 1+1/(1+2)+1/(1+2+3)+...+1/(1+2+3+...+100) 1+1/1+2+1/1+2+3+......+1/1+2+3+......+10 1/1+1/2+1/3+1/4+。。。。+1/N 是多少 (1-1/2^2)(1-1/3^2)....(1-1/10^2) (1-1/2^2)(1-1/3^2)....(1-1/10^2) s=1/1+1/1!+1/2！+1/3！+1/4！+....+1/25! (1-1/2)*(1+1/2)*(1-1/3)*(1+1/3)*……*（1-1/99）*（1+1/99）。 (1+1/2)*(1-1/2)(1+1/3)*(1-1/3)*.......(1+1/99))*(1-1/99)怎么做