初二下学期分式加减应用求解~已知:(1/m) - 1/(m+1) = 1/[m(m+1)] 请你化简:1/[(n+1)(n+2)] + 1/[n+2)(n+3)] +……+ 1/[(n+2006)(n+2007)] + 1/[(n+2007)(n+2008)]

来源:学生作业帮助网 编辑:作业帮 时间:2024/05/08 21:23:56
初二下学期分式加减应用求解~已知:(1/m) - 1/(m+1) = 1/[m(m+1)] 请你化简:1/[(n+1)(n+2)]  +  1/[n+2)(n+3)] +……+  1/[(n+2006)(n+2007)]  +  1/[(n+2007)(n+2008)]

初二下学期分式加减应用求解~已知:(1/m) - 1/(m+1) = 1/[m(m+1)] 请你化简:1/[(n+1)(n+2)] + 1/[n+2)(n+3)] +……+ 1/[(n+2006)(n+2007)] + 1/[(n+2007)(n+2008)]
初二下学期分式加减应用求解~
已知:(1/m) - 1/(m+1) = 1/[m(m+1)] 请你化简:1/[(n+1)(n+2)] + 1/[n+2)(n+3)] +……
+ 1/[(n+2006)(n+2007)] + 1/[(n+2007)(n+2008)]

初二下学期分式加减应用求解~已知:(1/m) - 1/(m+1) = 1/[m(m+1)] 请你化简:1/[(n+1)(n+2)] + 1/[n+2)(n+3)] +……+ 1/[(n+2006)(n+2007)] + 1/[(n+2007)(n+2008)]
1/[(n+1)(n+2)] + 1/[n+2)(n+3)] +……+ 1/[(n+2006)(n+2007)] + 1/[(n+2007)(n+2008)]
=1/(n+1)-1/(n+2)+1/(n+2)-1/(n+3)+······+1/(n+2007)-1/(n+2008)
=1/(n+1)-1/(n+2008)
=2007/(n+1)(n+2008)