(-1x1⼀2)+(-1⼀2x1⼀3)+(-1⼀3x1⼀4)+...+(-1⼀2004x1⼀2005)=???请说明过程

2025-04-16 01:36:41
推荐回答(2个)
回答1:

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

回答2:

(-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