作业帮1/X(X+1)+1/(X+1)(X+2)+1/(X+2)(X+3).+1/(X+99)(X+100)=
来源：学生作业帮助网 编辑：作业帮 时间：2024/04/30 02:29:32
1/X(X+1)+1/(X+1)(X+2)+1/(X+2)(X+3).+1/(X+99)(X+100)=
1/X(X+1)+1/(X+1)(X+2)+1/(X+2)(X+3).+1/(X+99)(X+100)=
1/X(X+1)+1/(X+1)(X+2)+1/(X+2)(X+3).+1/(X+99)(X+100)=
1/X(X+1)+1/(X+1)(X+2)+1/(X+2)(X+3).+1/(X+99)(X+100)
=1/X - 1/(X+1) + 1/(X+1) - 1/(X+2) + 1/(X+2) - 1/(X+3) + .+ 1/(X+99) - 1/(X+100)
= 1/x - 1/(x+100)
= 100/x(x+100)
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){...
全部展开
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)
收起
=100/X(X+100)
100/x(x+100);
1/x(x+1)=1/x-1/(x+1);
1/(x+1)(X+2)=1/(x+1)-1/(x+2);
.
.
.
1/(x+99)(x+100)=1/(x+99)-1/(x+100);
所以，结果消去中间相，得1/x-1/(x+100)=100/x(x+100)
100/X(X+100)
过程
1/X(X+1)+1/(X+1)(X+2)+1/(X+2)(X+3)......+1/(X+99)(X+100)
=[1/X-1/(X+1)]+[1/(X+1)-1/(X+2)]+[1/(X+2)-1/(X+3)]......+[1/(X+99)-1/(X+100)]
=1/X-1/(X+100)
=100/X(X+100)
把每项都分解下，
1/X(X+1)=1/X-1/(X+1);
1/(x+1)(x+2)=1/(x+1)-1/(x+2);
...
最后=1/x-1/(x+100)=100/X(X+100)
裂项相消
将每项都化成差的形式例
1/n(n+1)=1/n-1/(n+1)
你会发现可以消掉