1+1/(1+2)+1/(1+2+3)+……1/(1+2+3+……100)=?
来源:百度知道 编辑:UC知道 时间:2024/06/08 10:19:53
跪求!!!
要详细过程,没有的要结果也可以!急需啊!!!!!!!快!!!!!
要详细过程,没有的要结果也可以!急需啊!!!!!!!快!!!!!
任意一项1/(1+2+3+…n)=2/n(n+1)=2[1/n -1/(n+1)]
所以
1+1/(1+2)+1/(1+2+3)+……1/(1+2+3+……100)=2(1-1/2+1/2-1/3+1/3-1/4+1/4+……+1/99-1/100+1/100-1/101)=2(1-1/101)=200/101
公式编辑器无法用了^_^
1+2+3+……+n=n(n+1)/2
1/(1+2+3+……+n)=2/[n(n+1)]=2[1/n-1/(n+1)]
所以1+1/(1+2)+1/(1+2+3)+···1/(1+2+3+···+100)
=1+2*[(1/2-1/3)+(1/3-1/4)+……+(1/100-1/101)]
=1+2*(1/2-1/101)
=2-2/101
=200/101
1+1/(1+2)+1/(1+2+3)+……1/(1+2+3+……100)
=0.5/(1+6)*6/2/(1+100)*100/2
=0.5/21/5050
=0.00000471475......
(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+1/1+2+1/1+2+3+...+1/1+2+3...+2000
1+1/1+2+1/1+2+3.........+1/1+2+3.....100
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)
1+1/2+1+1/3+1+1/4+......+1/100=?
(1+1/2)(1+1/2^2)(1+1/2^4)(1+1/2^8)
(1-1/2^2)*(1-1/3^2)*(1-1/4^2).......(1-1/100^2)