奥数题1/1×2×3+1/2×3×4+……+1/98×99×100
答案:5 悬赏:50 手机版
解决时间 2021-04-01 02:54
- 提问者网友：浩歌待明月
- 2021-03-31 12:53
奥数题1/1×2×3+1/2×3×4+……+1/98×99×100
最佳答案
- 五星知识达人网友：狂恋
- 2021-03-31 13:59
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追问？过程？
=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追问？过程？
全部回答
- 1楼网友：三千妖杀
- 2021-03-31 18:34
裂项求和的方法求解前九十八项的和
- 2楼网友：何以畏孤独
- 2021-03-31 17:37
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
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
- 3楼网友：爱难随人意
- 2021-03-31 17:04
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
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
- 4楼网友：鸠书
- 2021-03-31 15:34
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
=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
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