作业帮奥数题1/1×2×3+1/2×3×4+……+1/98×99×100
来源：学生作业帮助网 编辑：作业帮 时间：2024/04/29 11:04:50
奥数题1/1×2×3+1/2×3×4+……+1/98×99×100
奥数题1/1×2×3+1/2×3×4+……+1/98×99×100
奥数题1/1×2×3+1/2×3×4+……+1/98×99×100
1/(1×2×3)+1/(2×3×4)+……+1/(98×99×100)
=1/2*[(1/(1*2)-1/(2*3)+1/(2*3)-1/(3*4)+……+1/(98*99-1/(99*100)]
=1/2*[1/2-1/9900]
=1/4-1/19800
=(4950-1)/19800
=4949/19800
1/(a-1)a(a+1)=1/2a[1/(a-1)-1/(a+1)]=1/2[1/(a-1)-2/a+1/(a+1)]
如1/98×99×100=1/2(1/98-2/99+1/100)
所以原式=1/2[(1-1+1/3)+(1/2-2/3+1/4)+...
=1/2(1-1+1/2+1/99-2/99+1/100)
=4949/19800
先做1/1×2+1/2×3+……+1/98×99这题，再做上边这题
首先得清楚1/(a-1)a=1/(a-1)-1/a;
则1/(a-1)a(a+1)=1/2a[1/(a-1)-1/(a+1)]=1/2[1/(a-1)-2/a+1/(a+1)]
原式=1/2[(1-1+1/3)+(1/2-2/3+1/4)+...(1/98-2/99+1/100)
=1/2(1-1+1/2+1/98-2/99+1/100)
=4949/19800
1/1×2×3+1/2×3×4+……+1/98×99×100
=1/2*（1/1*2-1/2*3)+1/2*(1/2*3-1/3*4)+......+1/2*(1/98*99-1/99*100)
=1/2*(1/1*2-1/99*100)
=1/4-1/19800
=4949/19800
裂项求和的方法求解前九十八项的和