查人人中国名人网辐射4年度版下载:求数列前n项和
来源：百度文库 编辑：查人人中国名人网 时间：2024/04/29 08:40:06
求
1/(1*2*3)+1/(2*3*4)+1/(3*4*5)+.....+1/(n*(n+1)*(n+2))
表达式~~~~
要证明哦
1/(1*2*3)+1/(2*3*4)+1/(3*4*5)+.....+1/(n*(n+1)*(n+2))
表达式~~~~
要证明哦
1/(n*(n+1)*(n+2))
=1/2*[1/(n*(n+1)) -1/((n+1)*(n+2))]
所以
1/(1*2*3)+1/(2*3*4)+1/(3*4*5)+.....+1/(n*(n+1)*(n+2))
=1/2*{1/(1*2)-1/(2*3)+1/(2*3)-1/(3*4)+...+1/[n*(n+1)]-1/[(n+1)*(n+2)]}
=1/2*{1/(1*2)-1/[(n+1)*(n+2)]}
=n(n+3)/[4(n+1)(n+2)]