k/(k+1)+1/(k+1)(k+2) =1-1/(k+1)+1/(k+1)-1/(k+2) =1-1/(k+2) =(k+1)/(k+2)k/(k+1)+1/(k+1)(k+2)……………………(1)=1-1/(k+1)+1/(k+1)-1/(k+2)…………(2)=1-1/(k+2)……………………………(3)=(k+1)/(k+2)…………………

来源:学生作业帮助网 编辑:作业帮 时间:2024/11/15 04:34:33
k/(k+1)+1/(k+1)(k+2) =1-1/(k+1)+1/(k+1)-1/(k+2) =1-1/(k+2) =(k+1)/(k+2)k/(k+1)+1/(k+1)(k+2)……………………(1)=1-1/(k+1)+1/(k+1)-1/(k+2)…………(2)=1-1/(k+2)……………………………(3)=(k+1)/(k+2)…………………
x)66@HSPUG126c5,Ê0|%h6Y1H밪I*ҧqE H ćD Ae|k]lg }loӎ 9#Y-Ovٌ+,Fā.|4XF̉P'6yvаs .*̀xOl>t=[ɎOz>\d

k/(k+1)+1/(k+1)(k+2) =1-1/(k+1)+1/(k+1)-1/(k+2) =1-1/(k+2) =(k+1)/(k+2)k/(k+1)+1/(k+1)(k+2)……………………(1)=1-1/(k+1)+1/(k+1)-1/(k+2)…………(2)=1-1/(k+2)……………………………(3)=(k+1)/(k+2)…………………
k/(k+1)+1/(k+1)(k+2) =1-1/(k+1)+1/(k+1)-1/(k+2) =1-1/(k+2) =(k+1)/(k+2)
k/(k+1)+1/(k+1)(k+2)……………………(1)
=1-1/(k+1)+1/(k+1)-1/(k+2)…………(2)
=1-1/(k+2)……………………………(3)
=(k+1)/(k+2)…………………………(4)
第(1)步是怎样得到第(2)步的啊

k/(k+1)+1/(k+1)(k+2) =1-1/(k+1)+1/(k+1)-1/(k+2) =1-1/(k+2) =(k+1)/(k+2)k/(k+1)+1/(k+1)(k+2)……………………(1)=1-1/(k+1)+1/(k+1)-1/(k+2)…………(2)=1-1/(k+2)……………………………(3)=(k+1)/(k+2)…………………
k/(k+1)+1/(k+1)(k+2)
=(k+1-1)/(k+1)+[(k+2)-(k+1)]/(k+1)(k+2)
=1-1/(k+1)+1/(k+1)-1/(k+2)

1/(k+1)(k+2)=1/k+1-1/k+2,建议楼主多看看书