一道数学题~!!!高手进啊 急,在线等

来源：百度知道 编辑：UC知道 时间：2024/09/12 22:51:45
1/X(X+2)+1/(X+2)(X+4)+...+1/(X+98)(X+100)
谁会帮忙啊!

1/X(X+2)+1/(X+2)(X+4)+...+1/(X+98)(X+100)
=1/2[1/x-1/(x+2)+1/(x+2)-1/(x+4)+1/(x+4)...+1/(x+98)-1/(x+100)]
=1/2[1/x-1/(x+100)
=50/x(x+100)

1/x(x+2)=1/2[1/x-1/(x+2)]
同理，原题等于
=1/2[1/x-1/(x+2)+1/(x+2)-1/(x+4).......-1/(x+98)+1/(x+98)-1/(x+100)]
=1/2[1/x-1/(x+100)]

解：1/X(X+2)=（1/x-1/(x+2））/2
1/(X+2)(X+4)=(1/(x+2）-1/(x+4）)/2
依此类推
1/X(X+2)+1/(X+2)(X+4)+...+1/(X+98)(X+100) =0.5*（1/x-1/(x+2)+1/(x+2)-1/(x+4)+1/(x+4）...+1/(x+98)-1/(x+100))
=0.5*(1/x-1/(x+100))

=(-1){1/2(1/(x+2)-1/x) + 1/2(1/(x+4)-1/(x+2))+...+1/2(1/(x+100)-1/(x+98))}
=(-1/2){-1/x+1/(x+100)}
=1/(2x)-1/(2x+200)

提2 加减相消