1/(1+2)+1/(1+2+3)+1/(1+2+3+4)+1/(1+2+3+4+5)+........+1/(1+2+3+4+......+100)=?
来源:百度知道 编辑:UC知道 时间:2024/06/26 04:18:21
1/(1+2+…+n)=1/〔(1+n)n÷2〕=2/(1+n)n=2×[1/n-1/(n+1)]
1/(1+2)+1/(1+2+3)+1/(1+2+3+4)+……+1/(1+2+3+……+100)
=2×(1/2-1/3)+2×(1/3-1/4)+2×(1/4-1/5)+……+2×(1/100-1/101)
=2×(1/2-1/3+1/3-1/4+1/4-1/5+……+1/100-1/101)
=2×(1/2-1/101)
=2×(99/202)
=99/101
求和 裂项。。
(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+1/2^2)(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+1/2+1+1/3+1+1/4+......+1/100=?