Question

1/x(x+1)(x+2)+1/(x+1)(x+2)(x+3)..........+1/(x+98)(x+99)(x+100)=?

Answer 1

1/x(x+1)(x+2)+1/(x+1)(x+2)(x+3)..........+1/(x+98)(x+99)(x+100)
=(1/2){[1/x(x+1) - 1/(x+1)(x+2)] + [1/(x+1)(x+2) - 1/(x+2)(x+3)]+............
+[1/(x+98)(x+99) - 1/(x+99)(x+100)]}
=(1/2){1/x(x+1) - 1/(x+99)(x+100)}
=[(x+99)(x+100) - x(x+1)]/2x(x+1)(x+99)(x+100)
=[198x+9900]/2x(x+1)(x+99)(x+100)
=[99x+4950]/x(x+1)(x+99)(x+100)
=99(x+50)/x(x+1)(x+99)(x+100)

Answer 2

1/x(x+1)(x+2)=1/x(x+2)-1/(x+1)(x+2)=1/2*1/x-1/(x+1)+1/2*1/(x+2)
1/(x+1)(x+2)(x+3)=1/(x+1)(x+3)-1/(x+2)(x+3)=1/2*1/(x+1)-1/(x+2)+1/2*1/(x+3)
1/(x+2)(x+3)(x+4)=1/(x+2)(x+4)-1/(x+3)(x+4)=1/2*1/(x+2)-1/(x+3)+1/2*1/(x+4)
。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。
1/(x+98)(x+99)(x+100)=1/(x+98)(x+100)-1/(x+99)(x+100)=1/2*1/(x+98)-1/(x+99)+1/2*1/(x+100)
左边相加=右边相加
原式=1/2*[1/x-1/(x+1)-1/(x+99)+1/(x+100)]

Answer 3

1/[(n-2)(n-1)n]
=1/2[1/(n-2)+1/n]-1/(n-1)
n=x+m
1/(x+m-2)(x+m-1)(x+m)
=1/2[1/(x+m-2)+1/(x+m)]-1/(x+m-1)
1/x(x+1)(x+2)+1/(x+1)(x+2)(x+3)..........+1/(x+98)(x+99)(x+100)
=1/2[1/x+1/(x+2)+1/(x+1)+1/(x+3)+1/(x+2)/1/(x+4)......1/(x+98)+1/(x+97)+1/(x+99)+1/(x+98)+1/(x+100)]-1/(x+1)-1/(x+2)-...-1/(x+99)
=1/2*[1/x-1/(x+1)-1/(x+99)+1/(x+100)]
=99(x+50)/[x(x+1)(x+99)(x+100)]

Answer 4

1/x(x+1)(x+2)+1/(x+1)(x+2)(x+3)..........+1/(x+98)(x+99)(x+100)
=[1/x(x+1)-1/(x+1)(x+2)]/2+[1/(x+1)(x+2)-1/(x+2)(x+3)]/2+...+[1/(x+98)(x+99)-1/(x+99)(x+100)]/2
=[1/x-2/(x+1)+1/(x-2)]/2+[1/(x+1)-2/(x+2)+1/(x+3)]/2+...+[1/(x+98)-2/(x+99)+1/(x+100)]/2
=[1/x-1/(x+1)-1/(x+99)+1/(x+100)]/2

Answer 5

其实可设：1/x(x+1)(x+2)=a/x+b/(x+1)+c/(x+2)，
通过通分，代数式对应的系数相等，得到：
a+b+c=0，3a+2b+c=0，2a=1，
解得：a=1/2，b=-1，c=1/2。
所以1/x(x+1)(x+2)=1/2*[1/x-2/(x+1)+1/(x+2)]。
同理可推：
1/(x+1)(x+2)(x+3)=1/2*[1/(x+1)-2/(x+2)+1/(x+3)]，
.....................................................
1/(x+98)(x+99)(x+100)=1/2*[1/(x+98)-2/(x+99)+1/(x+100)]，
所以
原式=1/2*{[1/x-2/(x+1)+1/(x+2)]+[1/(x+1)-2/(x+2)+1/(x+3)]+....+[1/(x+98)-2/(x+99)+1/(x+100)]}
=1/2*[1/x-1/(x+1)-1/(x+99)+1/(x+100)]
=99(x+50)/x(x+1)(x+99)(x+100)。