Question

七年级有理数问题

请先阅读下列一组内容，然后解答问题：
1/1x2+1/2x3+1/3x4...+1/9x10=（1-1/2）+(1/2-1/3)+(1/3-1/4)+...+(1/9-1/10)=1+1/2-1/3+1/3+1/4+...+1/9-1/10=9/10

问题：
计算：①1/1x2+1/2x3+1/3x4+...+1/2004x2005=

②1/1x3+1/3x5+1/5x7+...+1/49x51=

Answer 1

：①1/1x2+1/2x3+1/3x4+...+1/2004x2005
=1-1/2+1/2-1/3+……+1/2004-1/2005
=1-1/2005
=2004/2005

②1/1x3+1/3x5+1/5x7+...+1/49x51
=1/2×（1-1/3+1/3-1/5+……+1/49-1/51）
=1/2×（1-1/51）
=1/2×50/51
=25/51

Answer 2

①1/1x2+1/2x3+1/3x4+...+1/2004x2005=（1-1/2）+(1/2-1/3)+(1/3-1/4)+...+(1/2004-1/2005)=
1-1/2005=2004/2005
②1/1x3+1/3x5+1/5x7+...+1/49x51=1/2*[(1-1/3)+(1/3-1/5)+(1/5-1/7)+...+(1/49-1/51)]=2*(1-1/51)25/51

Answer 3

1/1x2+1/2x3+1/3x4+...+1/2004x2005=（1-1/2）+(1/2-1/3)+(1/3-1/4)+...+(1/2004-1/2005)=
1-1/2005=2004/2005
1/1x3+1/3x5+1/5x7+...+1/49x51=1/2*[(1-1/3)+(1/3-1/5)+(1/5-1/7)+...+(1/49-1/51)]=2*(1-1/51)25/51 裂项相消高中会学的