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!
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