1/1x2x3+1/2x3x4+1/3x4x5+.+1/n(n+1)(n+2)
来源:学生作业帮 编辑:作业帮 分类:数学作业 时间:2024/11/09 03:36:34
1/1x2x3+1/2x3x4+1/3x4x5+.+1/n(n+1)(n+2)
Sn=1/[n(n+1)(n+2)]=(1/2){1/[n)n+1)]-1/[(n+1)(n+2)]}
=(1/2)[1/n-1/(n+1)-1/(n+1)+1/(n+2)]
=(1/2)[1/n-2/(n+1)+1/(n+2)]
=1/1x2x3+1/2x3x4+1/3x4x5+...+1x/n(n+1)(n+2)
=(1/2)[1/1-2/2+1/3+1/2-2/3+1/4+1/3-2/4+1/5+/4-2/5+1/6
+.+1/n-2/(n+1)+1/(n+2)]
=(1/2)[1-1/2-1/(n+1)+1/(n+2)]
=(n^2+3n)/[4(n+1)(n+2)]
=(1/2)[1/n-1/(n+1)-1/(n+1)+1/(n+2)]
=(1/2)[1/n-2/(n+1)+1/(n+2)]
=1/1x2x3+1/2x3x4+1/3x4x5+...+1x/n(n+1)(n+2)
=(1/2)[1/1-2/2+1/3+1/2-2/3+1/4+1/3-2/4+1/5+/4-2/5+1/6
+.+1/n-2/(n+1)+1/(n+2)]
=(1/2)[1-1/2-1/(n+1)+1/(n+2)]
=(n^2+3n)/[4(n+1)(n+2)]
1x2x3+2x3x4+3x4x5+4x5x6+...+n(n+1)(n+2)=
1/(1x2x3)+1/(2x3x4)+1/(3x4x5)+.1/(nx(n+1)x(n+2)=?
求和1x2x3+2x3x4+...+n(n+1)(n+2)
1/1x2x3+1/2x3x4+1/3x4x5
1x2X3+2x3X4+3x4X5+…+7X8X9=?
1x2x3+2x3x4+3x4x5+…+8x9x10
1x2x3+2x3x4+3x4x5+...+7x8x9=,
1x2x3+2x3x4+3x4x5+.+10x11x12
1/1x2x3+1/2x3x4+1/3x4x5+1/4x5x6+.+1/48x49x50=
1/1x2x3+1/2x3x4+1/3x4x5+.+1/11x12x13=
1/1x2x3+1/2x3x4+1/3x4x5+.+1/9x10x11=
请1/1x2x3+1/2x3x4+1/3x4x5+1/4x5x6=