1/1*2+1/2*3+1/3*4+.......+1/2004*2005的和是多少?
来源:百度知道 编辑:UC知道 时间:2024/05/13 19:58:26
1/1*2+1/2*3+1/3*4+.......+1/2004*2005
=(1-1/2)+(1/2-1/3)+(1/3-1/4)+……+(1/2004-1/2005)
=1-1/2+1/2-1/3+1/3-……+1/2004-1/2005
=1-1/2005
=2004/2005
2004/2005
每一项变成1/1-1/2+1/2-1/3+...+1/2004-1/2005
除了第一项和最后一项外 中间的都抵消了
所以最后答案1-1/2005=2004/2005
每一项变成1/1-1/2+1/2-1/3+...+1/2004-1/2005
然后抵消中间各项为:1+0+0+0……-1/2005
最后结果为1-1/2005=2004/2005
解:原式=(1-1/2)+(1/2-1/3)+(1/3-1/4)+……+(1/2004-1/2005)
=1-1/2+1/2-1/3+1/3-1/4+……1/2004-1/2005
=1-1/2005
=2004/2005
原式=(1-1/2)+(1/2-1/3)+(1/3-1/4)+......+(1/2004-1/2005)+2005
=1-1/2+1/2-1/3+1/3-1/4+......1/2004-1/2005+2005
=1-1/2005+2005
=2005+2004/2005
(1/2005-1)(1/2004-1)........(1/3-1)(1/2-1)
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)×
(1-1/2)(1-1/3)(1-1/4)(1-1/5).....(1-1/1000)
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