数学题目，分式的化简：1/a-1 + 1/(a-1)(a-2) + 1/(a-2)(a-3) +.....+ 1/(a-99)(a-100)

Question

最好要有解答过程，清晰点，我读初二！
一般的我是明白的！ 辛苦各位了！

Answer 1

1=(a-1)-(a-2)
所以1/(a-1)(a-2)=[(a-1)-(a-2)]/(a-1)(a-2)=(a-1)/(a-1)(a-2)-(a-2)/(a-1)(a-2)=1/(a-2)-1/(a-1)

同理，1=(a-2)-(a-3),得到1/(a-2)(a-3)=1/(a-3)/1/(a-2)
后面以此类推
所以原式=1/(a-1)+1/(a-2)-1/(a-1)+1/(a-3)/1/(a-2)+……+1/(a-100)-1/(a-99)
=1/(a-100)

Answer 2

解：
1/(a-1) + 1/(a-1)(a-2) + 1/(a-2)(a-3) +.....+ 1/(a-99)(a-100)
＝1/(a-1) + [1/(a-2)-1/(a-1)]+ [1/(a-3))-1/(a-2)]+.....+ [1/(a-100)-1/(a-99)]
＝1/(a-100)