Question

一道初二数学题,不是很难的,帮帮忙吧!!⊙o⊙

提示:a/b就相当于b分之a.
计算:
1/a(a+1)+1/(a+1)(a+2)+1/(a+2)(a+3)+......+1/(a+2003)(a+2004)

Answer 1

1/a(a+1)+1/(a+1)(a+2)+1/(a+2)(a+3)+......+1/(a+2003)(a+2004)
=(1/a-1/a+1)+(1/a+1-1/a+2)+......+(1/a+2002-1/a+2003)+(1/a+2003-1/a+2004)
等于-1/a+1+1/a+1可以削去 就是后一项和下一项和为0
所以此题化简为1-1/a+2004
通分后答案为2004/a(a+2004)

Answer 2

把他们都给拆开~
然后就这样了
1/a-1/（a+1）+1/（a+1）-1/（a+2）+...+1/（a+2003）-1/（a+2004）
最后就是1/a-1/（a+2004）

Answer 3

=1/a-1/(1+a)+1/(1+a)-1/(2+a)....+1/(a+2003)-1/(a+2004)
=1/a-1/(a+2004)
=2004/a(a+2004)

Answer 4

=1/a-1/(a+1)+1/(a+1)-1/(a+2).......+1/(a+2003)-1/(a+2004)=1/a-1/(a+2004)=2004/a(a+2004)

Answer 5

化分母！
((a+1)-a)+...+1/((a+2004)-(a+2003))
=2004/a(a+2004)