(1/2+1/3+1/4+...+1/20)+(2/3+2/4+2/5...+2/20
来源:百度知道 编辑:UC知道 时间:2024/05/14 18:52:53
(1/2+1/3+1/4+...+1/20)+(2/3+2/4+2/5...+2/20)+(3/4+3/5+...+3/20)+...+(18/19+18/20)+19/20=
(1/2+1/3+1/4+...+1/20)+(2/3+2/4+2/5...+2/20)+(3/4+3/5+...+3/20)+...+(18/19+18/20)+19/20
[1+2+3+...+(n-1)]/n=n(n-1)/2n=(n-1)/2
(1/2+1/3+1/4+...+1/20)+(2/3+2/4+2/5+...+2/20)+(3/4+3/5+...+3/20)+...+(18/19+18/20)+19/20
=1/2+(1/3+2/3)+(1/4+2/4+3/4)+...+(1/20+2/20+3/20+...+19/20)
=(2-1)/2+(3-1)/2+(4-1)/2+...+(20-1)/2
=(1+2+3+...+20)/2
=20*21/4
=5*21
=105
等差数列,1/2为首项,1/2为等差,19项。和为[1/2+1/2+(1/2)*18]*19/2=95
(1/2005-1)(1/2004-1)........(1/3-1)(1/2-1)
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+----+100)
1+1/2+1/3+.....+1/n
1+1/2+1/3+...+1/100
1-1/2+1/3-.....-1/10
(1+1/2+1/3+1/4)×
1+1/1+2+1/1+2+3+...+1/1+2+3...+2000
1*(1/1+2)*(1/1+2+3)*~~~*(1/1+2+~~~2005)=?
(1-1/2)(1-1/3)(1-1/4)(1-1/5).....(1-1/1000)