1+1/1*2+1/2*3+1/3*4…+1/2007*2008 1+1/1*2+1/2*3+1/3*4…+1/2007*2008
来源:百度知道 编辑:UC知道 时间:2024/06/25 05:13:51
解:
原题
=1+(2-1)/(2*1) + (3-2)/(2*3) +(4-3)/(3*4)+....+(2008-2007)/(2007*2008)
=1+(1-1/2)+(1/2-1/3)+(1/3-1/4)+....+(1/2007-1/2008)
=1+1-1/2008
=4015/2008。
=1+1-1/2+1/2-1/3+..........+1/2007-1/2008=4015/2008
解:
原式=(2-1)/(2*1) + (3-2)/(2*3) +(4-3)/(3*4)+....+(2008-2007)/(2007*2008)
=(1-1/2)+(1/2-1/3)+(1/3-1/4)+....+(1/2007-1/2008)
=1-1/2008
=2007/2008。
解题过程:
原式=1+(1-1/2+1/2-1/3+1/3-1/4+…-1/2007+1/2007-1/2008)
=1+(1-1/2008)
=2-1/2008
=4015/2008
本题用的是裂项相消法,1/1*2=1-1/2,同理,1/2*3=1/2-1/3,以此类推,就会消除中间各项,只剩首尾两项。
(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)