1/1*2+1/2+3+···1/2002*2003+1/2003*2004+1/2004*2005

问题描述:

1/1*2+1/2+3+···1/2002*2003+1/2003*2004+1/2004*2005

1/1*2+1/2*3+···1/2002*2003+1/2003*2004+1/2004*2005
=( 1-1/2)+(1/2-1/3)+(1/3-1/4)+.....+(1/2003-1/2004)+(1/2004-1/2005)
= 1-1/2005
=2004/2005

根据1/n*(n+1)=1/n-1/(n+1),
1/1*2+1/2+3+···1/2002*2003+1/2003*2004+1/2004*2005
=1-1/2+1/2-1/3+....+1/2004-1/2005
=1-1/2005
=2004/2005

1/1*2+1/2+3+···1/2002*2003+1/2003*2004+1/2004*2005
=(1-1/2)+(1/2-1/3)+(1/3-1/4)+...+(1/2004-1/2005)
=1-1/2005
=2004/2005