3/(1!+2!+3!)+4/(2!+3!+4!)+...+2000/(1998!+1999!+2000!)
问题描述：
3/(1!+2!+3!)+4/(2!+3!+4!)+...+2000/(1998!+1999!+2000!)
这个题解了很久也没解出来,接不出来爸爸不让吃饭!十万火急!
答
先看通项：
(n+2)/[n!+(n+1)!+(n+2)!]
=(n+2)/[n!(1+n+1)+(n+2)(n+1)!]
=(n+2)/{(n+2)[n!+(n+1)!]}
=1/[n!+(n+1)!]
=1/[(n+2)n!]
=(n+1)/[(n+1)(n+2)n!]
=[(n+2)-1]/(n+2)!
=1/(n+1)!-1/(n+2)!
∴
3/(1!+2!+3!)=1/2!-1/3!
4/(2!+3!+4!)=1/3!-1/4!
……
2000/(1998!+1999!+2000!)=1/1999!-1/2000!
以上各式相加,右面中间项交叉相消
∴原式=1/2!-1/2000!=1/2-1/2000!