1/(1*2)+1/(2*3)+1/(3*4)+1/(4*5)+......+1/(39*40)
来源:百度知道 编辑:UC知道 时间:2024/06/02 04:41:36
原式=1-1/2+1/2-1/3+……+1/39-1/40=1-1/40=39/40
1/(1*2)+1/(2*3)+1/(3*4)+1/(4*5)+......+1/(39*40)
=1-1/2+1/2-1/3+1/3-1/4+....-1/39+1/39-1/40
=1-1/40
39/40
(1-1/2)+(1/2-1/3)+…(1/39-1/40)=1-1/40=39/40
1/(1*2)=1-1/2 同理可得:
1/(2*3)=1/2-1/3…………
所以:
原式=1-1/2+1/2-1/3+1/3-1/4……+1/38-1/39+1/39-1/40=1-1/40=39/40
此所谓错位抵消!!
1/2-1/2=?
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)
3/2=2+1/1*2=1/1+1/2
(1/2005-1)(1/2004-1)........(1/3-1)(1/2-1)