1/1×2×3×4×5 + 1/2×3×4×5×6 + 1/3×4×5×6×7 + 1/4×5...
发布网友
发布时间：2024-10-03 17:28
共1个回答
热心网友
时间：2024-10-06 18:29
解：
1/[n(n+1)(n+2)(n+3)(n+4)]
=¼×[(n+4)-n]/[n(n+1)(n+2)(n+3)(n+4)]
=¼×{ 1/[n(n+1)(n+2)(n+3)] -1/[(n+1)(n+2)(n+3)(n+4)]}
1/(1×2×3×4×5)+ 1/(2×3×4×5×6)+1/(3×4×5×6×7)+ 1/(4×5×6×7×8)
=¼×[1/(1×2×3×4)- 1/(2×3×4×5)+ 1/(2×3×4×5)- 1/(3×4×5×6)+1/(3×4×5×6)- 1/(3×5×6×7)+ 1/(4×5×6×7)- 1/(5×6×7×8)]
=¼×[1/(1×2×3×4)-1/(5×6×7×8)]
=¼×(1/24)[1- 1/(5×2×7)]
=¼×(1/24)([1- 1/70)
=¼×(1/24)(69)/70
=¼×(1/8)(23)/70
=23/2240
