(-1x1/2)+(-1/2x1/3)+(-1/3x1/4)+...+(-1/2004x1/2005)=?
问题描述：
(-1x1/2)+(-1/2x1/3)+(-1/3x1/4)+...+(-1/2004x1/2005)=?
答
解:原式=-[1/2+1/6+1/12+···+1/(2004·2005)] =-(1/2+1/2-1/3+1/3-1/4+1/4+···+1/2004-1/2005) =-(1-1/2005)= -2004/2005
答
(-1×1/2)＋(-1/2×1/3)＋(-1/3×1/4)＋……＋(-1/2004×1/2005)
＝-1/(1×2)＋[-1/(2×3)]＋[-1/(3×4)]＋……＋[-1/(2004×2005)]
＝-(1/1－1/2)＋[-(1/2－1/3)]＋[-(1/3－1/4)]＋……＋[-(1/2004－1/2005)]
＝-1＋1/2－1/2＋1/3－1/3＋1/4－……－1/2004＋1/2005
＝-1＋1/2005
＝-2004/2005