初2一道数学题,速求解

1/a(a+1)+1/(a+1)(a+2)+1/(a+2)(a+3).........+1(a+2007)(a+2008)

第一项1/a(a+1)=1/a-1/(a+1)
第二项1/(a+1)(a+2)=1/(a+1)-1/(a+2)
第三项1/(a+2)(a+3)=1/(a+2)-1/(a+3)
.........
第2008项1/(a+2007)(a+2008)=1/(a+2007)-1/(a+2008)

等式两边相加得
1/a(a+1)+1/(a+1)(a+2)+1/(a+2)(a+3).........+1(a+2007)(a+2008)
=1/a-1/(a+1)+1/(a+1)-1/(a+2)+1/(a+2)-1/(a+3).........+1/(a+2007)-1/(a+2008)
=1/a+1/(a+2008)
=2008/a(a+2008)

=[1/a-1/(a+1)]+[1/(a+1)-1/(a+2)]+…………+[1/(a+2007)-1/(a+2008)]
=1/a-1/(a+2008)
=2008/a(a+2008)

拆解
如:1/a(a+1)＝1/a - 1/(a+1)
1/(a+1)(a+2)=1/(a+1) - 1/(a+2)
类推
则原式
=1/a - 1/(a+2008)
= 2008/a(a+2008)

裂项！1/a(a+1)=1/a-1/(a+1) 依次类推 最终得1/a-1/(a+2008)

裂项求和的问题！将1/a(a+1)变成1/a-1/(a+1)这是做这类形题的规律！如此类推！1/(a+1)(a+2)=1/(a+1)-1/(a+2)写几项出来发现前后可以相消！搞定了！