1/[n(n+1)]+1/[(n+1)(n+2)]+1/[(n+2)(n+3)]+1/[(n+3)(n+4)]+[(n+4)(n+_数学解答

1/[n(n+1)]+1/[(n+1)(n+2)]+1/[(n+2)(n+3)]+1/[(n+3)(n+4)]+[(n+4)(n+5)}怎么算谢谢了,

人气：457 ℃　时间：2020-06-04 00:33:11

解答

1/[n(n+1)]=1/n-1/(n+1) 因此,上式=(1/n-1/(n+1))+(1/(n+1)-1/(n+2))+...+(1/(n+4)-1/(n+5)) =1/n-1/(n+5) =5/[n(n+5)]