初一数学计算题（寻找规律）[1]

来源：百度知道 编辑：UC知道 时间：2024/05/16 19:37:51
(a=2 b=1)
求1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+…+1/(a+2002)(b+2002)的值
谢谢！详解！

1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+…+1/(a+2002)(b+2002)
=1/2*1+1/(2+1)(1+1)+1/(2+2)(1+2)+......+1/(2+2002)(1+2002)
=1/1*2+1/2*3+1/3*4+......+1/2003*2004
=(1-1/2)+(1/2-1/3)+(1/3-1/4)+......+(1/2003-1/2004)
=1-1/2+1/2-1/3+1/3-1/4+......+1/2003-1/2004
=1-1/2004
=2003/2004

1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+…+1/(a+2002)(b+2002)
=1/2*1+1/(2+1)(1+1)+1/(2+2)(1+2)+......+1/(2+2002)(1+2002)
=1/1*2+1/2*3+1/3*4+......+1/2003*2004
=(1-1/2)+(1/2-1/3)+(1/3-1/4)+......+(1/2003-1/2004)
=1-1/2+1/2-1/3+1/3-1/4+......+1/2003-1/2004
=1-1/2004
=2003/2004
这样行吧？

很想帮助你，但是不会