作业帮证明0.0999＜1/10"+1/11"+1/12"+.+1/1000"＜0.111证明：0.0999＜[1/(10x10)+1/(11x11)+1/(12x12)+.....+1(1000x1000)]＜0.111
来源：学生作业帮助网 编辑：作业帮 时间：2024/05/05 06:57:22
证明0.0999＜1/10"+1/11"+1/12"+.+1/1000"＜0.111证明：0.0999＜[1/(10x10)+1/(11x11)+1/(12x12)+.....+1(1000x1000)]＜0.111
证明0.0999＜1/10"+1/11"+1/12"+.+1/1000"＜0.111
证明：0.0999＜[1/(10x10)+1/(11x11)+1/(12x12)+.....+1(1000x1000)]＜0.111
证明0.0999＜1/10"+1/11"+1/12"+.+1/1000"＜0.111证明：0.0999＜[1/(10x10)+1/(11x11)+1/(12x12)+.....+1(1000x1000)]＜0.111
[1/(10x10)+1/(11x11)+1/(12x12)+.+1(1000x1000)]>[1/(10x11)+1/(11x12)+1/(12x13)+.+1(1000x1001)]=(1/10)-(1/11)+(1/11)-(1/12)+.+(1/1000)-(1/1001)=
(1/10)-(1/1001)=0.099
[1/(10x10)+1/(11x11)+1/(12x12)+.+1(1000x1000)]
"什么意思啊
应该利用以下两式:
1/[(n(n+1)] < 1/n^2 < 1/[(n-1)n]
1/[n(n+1)] = 1/n - 1/(n+1)
则原式 > 1/(10x11)+1/(11x12)+1/(12x13)+.....+1/(1000x1001)
= 1/10 - 1/1001 = 0.099000999000999...
原式 < 1/(9x...
全部展开
应该利用以下两式:
1/[(n(n+1)] < 1/n^2 < 1/[(n-1)n]
1/[n(n+1)] = 1/n - 1/(n+1)
则原式 > 1/(10x11)+1/(11x12)+1/(12x13)+.....+1/(1000x1001)
= 1/10 - 1/1001 = 0.099000999000999...
原式 < 1/(9x10)+1/(10x11)+1/(11x12)+.....+1/(999x1000)
= 1/9 - 1/1000 = 0.110111111111..< 0.111
以上可以证明原式< 0.111,但还不能证明原式 > 0.0999.所以要考虑更精确的.先把头尾两项去掉,求出剩余的取值范围,再把头尾两项加回.
1/(11x11)+1/(12x12)+....+1/(999x999) > 1/(11x12)+1/(12x13)+....+1/(999x1000) = 1/11 - 1/1000 = 989 /11000
则有: 1/(11x11)+1/(12x12)+....+1/(999x999) > 989 /11000
把头尾两项加上得:
原式 > 989 /11000 + 1/(10x10) + 1(1000x1000) =
0.099910090909>0.0999
所以, 0.0999＜[1/(10x10)+1/(11x11)+1/(12x12)+.....+1(1000x1000)]＜0.111
收起
这是一个很简单的数学公式而已啊