Question

一道初一找规律的数学题，下面是题，我主要要的是讲解啊！

a=2，b=1
1/ab+1/（a+1)(b+1)+1/（a+2)(b+2)+·····(a+2005)(b+2005）的值

Answer 1

解法如下：
1/(a+2004)(b+2004)=1/2005-1/2006
1/(a+2005)(b+2005)=1/2006-1/2007
以上各式相加得
1-1/2+1/2-1/3+1/3-1/4…………+1/2005-1/2006+1/2006-1/2007
=1-1/2007
=2006/2007
希望能帮到你

Answer 2

因为a=2，b=1
所以1/ab+1/（a+1)(b+1)+1/（a+2)(b+2)+•……+1/(a+2005)(b+2005）
=1/(2×1)+1/(3×2)+1/(4×3)+……+1/(2007×2006)
=(1-1/2)+(1/2-1/3)+(1/3-1/4)+……+(1/2006-1/2007)
=1+(-1/2+1/2)+(-1/3+1/3)+(-1/4+1/4)……+(-1/2006+1/2006)-1/2007
=1-1/2007=2006/2007

Answer 3

1/ab = 1/(2*1)=1/1-1/2
1/(a+1)(b+1)=1/(3*2)=1/2-1/3
1/(a+2)(b+2)=1/(4*3)=1/3-1/4
……………………
1/(a+2004)(b+2004)=1/2005-1/2006
1/(a+2005)(b+2005)=1/2006-1/2007
以上各式相加得
1-1/2+1/2-1/3+1/3-1/4…………+1/2005-1/2006+1/2006-1/2007
=1-1/2007
=2006/2007

Answer 4

1/ab+1/（a+1)(b+1)+1/（a+2)(b+2)+·····(a+2005)(b+2005）
=1/2*1+1/3*2+1/4*3+……1/2007*2006
=1-1/2+1/2-1/3+1/3-1/4+……1/2006-1/2007
=1-1/2007
=2006/2007