化简 1／X×（X+2）＋1／（X+2）×（X+4）+…+1／（X+2006）×（X+2008）
问题描述：
化简 1／X×（X+2）＋1／（X+2）×（X+4）+…+1／（X+2006）×（X+2008）
答
1/2*[1/x-1/(x+2)+1/(x+2)-1/(x+4)+.....+1/(x+2006)-1/(x+2008)]
=1/2*[1/x-1/(x+2008)]
=2007/[2*x*(x+2008)]
答
原式
=1/2×[1/x-1/(x+2)]+1/2×[1/(x+2)-1/(x+4)]+.+1/2×[1/(x+2006)-1/(x+2008)]
=1/2×[1/x-1/(x+2)+1/(x+2)-1/(x+4)+.+1/(x+2006)-1/(x+2008)]
=1/2×[1/x-1/(x+2008)]
=1/2×[(x+2008)-x]/[x(x+2008)]
=1/2×2008/[x(x+2008)]
=1004/[x(x+2008)]
=1004/(x^2+2008x)
这种方法在数学中叫做‘裂项相消法’.