中科新闻网谷歌邮箱在中国怎么用:数学初一问题
来源：百度文库 编辑：中科新闻网 时间：2024/04/27 19:29:05
计算1/[x+1)(x+2)]+1/[(x+2)(x+3)]+1/[(x+3)(x+4)]+……+1/[(x+2003)(x+2004)]=
=[1/(x+1)-1/(x+2)]+[1/(x+2)-1/(x+3)]+[1/(x+3)-1/(x+4)]……+[1/（x+2003)-1/(x+2004)]
=1/(x+1)-1/(x+2004)
=2003/[(x+1)(x+2004)]
裂项相消
=1/(x+1)-1/(x+2)+1/(x+2)-1/(x+3)+...+1/(x+2003)-1/(x+2004)
=1/(x+1)-1/(x+2004)
=2003/[(x+1)(x+2004)]