(1-1/2*2)(1-1/3*3)(1-1/4*4)(1-1/5*5)......(1-1/2007*2007)的答案及过程是怎样的?
来源:百度知道 编辑:UC知道 时间:2024/09/14 17:36:25
1-1/2*2=(1/2)*(3/2)
1-1/3*3=(2/3)*(4/3)
(1-1/2*2)(1-1/3*3)(1-1/4*4)(1-1/5*5)......(1-1/2007*2007)=(1/2)(3/2)……2008/2007=2008/2007
我不是很清楚楼上的算法,可是我记得好象是先乘除后加减吧?那(1-1/2*2)是不是应该等于0啊?
1-1/2^2=(1/2)*(3/2) 1-1/2^2=(2^2-1/)2^2=3*1/2^2=(3/2)*(1/2)
1-1/3^3=(2/3)*(4/3)
...
1-1/2007^2007=(2006/2007)*(2008/2007)
(1-1/2^2)(1-1/3^3)(1-1/4^4)(1-1/5^5)......(1-1/2007^2007)
=(1/2)(3/2)……2008/2007
=1/2*2008/2007
=1004/2007
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+1/2^2)(1+1/2^4)(1+1/2^8)
(1-1/2^2)*(1-1/3^2)*(1-1/4^2).......(1-1/100^2)
1/2-1/2=?
3/2=2+1/1*2=1/1+1/2
(1/2005-1)(1/2004-1)........(1/3-1)(1/2-1)