1/(1*3)+1/(2*4)+1/(3*5)+......+1/(101*103)
来源:百度知道 编辑:UC知道 时间:2024/06/14 01:54:52
1/[n*(n+2)]= [1/n - 1/(n+2)]/2
1/(1*3)+1/(2*4)+1/(3*5)+......+1/(101*103)
=[1/(1*3)+1/(3*5)+...+1/(101*103)]+[1/(2*4)+1/(4*6)+...+1/(100*102)]
=(1-1/3+1/3-1/5++....+1/101-1/103)/2 + (1/2-1/4+1/4-1/6+...+1/100-1/102)/2
=(1-1/103)/2+(1/2-1/102)/2
=51/103+25/102
1/(1*3)+1/(2*4)+1/(3*5)+……+1/(101*103)
=1/2*((1-1/3)+(1/2-1/4)+(1/3-1/5)+……+(1/101-1/103))
=1/2*((1-1/3+1/2-1/4+1/3-1/5+1/4-1/6+1/5-1/7+……+1/101-1/103))
=1/2(1+1/2-1/103)
=257/206
(1/2005-1)(1/2004-1)........(1/3-1)(1/2-1)
(1-1/100)(1-1/99)(1-1/98)......(1-1/3
(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+1/2+1+1/3+1+1/4+......+1/100=?
1+1/2+1/3+.....+1/n
1+1/2+1/3+...+1/100
1+1/3+1/6+........+1/55
1-1/2+1/3-.....-1/10