(1-1/2-1/3-...-1/2001)(1/2+1/3+...1/2002)-(1-1/2-1/3-...1/2002)(1/2+1/3+...+1/2001)

来源：百度知道编辑：UC知道时间：2024/05/12 16:32:43

1-1/2-1/3-...-1/2001)*(1/2+1/3+...1/2002)-(1-1/2-1/3-...1/2002)*(1/2+1/3+...+1/2001)

设（1/2+1/3+...1/2001）为X，（1/2+1/3+...1/2002）为Y，原式则可以写为：
（1-X）*Y-（1-Y）*X
=Y-XY-X+XY
=Y-X
=1/2002

(1-1/2-1/3-...-1/2001)*(1/2+1/3+...1/2002)-(1-1/2-1/3-...1/2002)*(1/2+1/3+...+1/2001)
=(1-1/2-1/3..-1/2001)(1/2+1/3+...1/2001)+(1-1/2-1/3..1/2001)*1/2002-[(1-1/2-1/3...-1/2001)(1/2+1/3+...1/2001)-1/2002*(1/2+1/3+...1/2001)
=(1-1/2-1/3..-1/2001)(1/2+1/3+...1/2001)+(1-1/2-1/3..1/2001)*1/2002-(1-1/2-1/3...-1/2001)(1/2+1/3+...1/2001)+1/2002*(1/2+1/3+...1/2001)
=1/2002*(1-1/2-1/3-1/2001+1/2+1/3+...1/2001)
=1/2002*1
=1/2002

(1+1/2+1/3+1/4)× (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) 1/1+1/2+1/3+1/4+。。。。+1/N 是多少 1/1+1/2+1/3+1/4+......1/2002＝？ 1-1/2+1/3-1/4+........1/99-1/100 求和Sn=1+(1+1/2)+(1+1/2+1/4)+....+[1+1/2+1/4.....+1/2^(n-1)] 数列 1+(1+1/2)+(1+1/2+1/4)+..............=?