高考问答疯狂联盟手游下载:请问这个题该怎么做:[1/(1+1)!]+[2/(2+1)!]+[3/(3+1)!]+......+[n/(n+1)!]=1-[1/(n+1)!]
来源：百度文库 编辑：高考问答 时间：2024/05/09 02:21:31
[1/(1+1)!]+[2/(2+1)!]+[3/(3+1)!]+[4/(4+1)!]……+[n/(n+1)!]
=[1-1/(1+1)!]+[2/(2+1)!]+[3/(3+1)!]+[4/(4+1)!]……+[n/(n+1)!]
=1-[1/(1+1)!-2/(2+1)!]+[3/(3+1)!]+[4/(4+1)!]……+[n/(n+1)!]
=1-[(2+1)/(2+1)!-2/(2+1)!]+[3/(3+1)!]+[4/(4+1)!]……+[n/(n+1)!]
=1-[1/(2+1)!]+[3/(3+1)!]+[4/(4+1)!]……+[n/(n+1)!]
=1-[1/(2+1)!-3/(3+1)!]+[4/(4+1)!]……+[n/(n+1)!]
=1-[(3+1)/(3+1)!-3/(3+1)!]+[4/(4+1)!]……+[n/(n+1)!]
=1-[1/(3+1)!]+[4/(4+1)!]……+[n/(n+1)!]
…………
=1-[1/(n+1)!]
这个式子等于1-一大堆东西呢