Chứng minh rằng : \(\dfrac{1}{6}< \dfrac{1}{5^2}+\dfrac{1}{6^2}+\dfrac{1}{7^2}+....+\dfrac{1}{100^2}< \dfrac{1}{4}\)
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NP
2
6 tháng 9 2021
\(\dfrac{1}{5^2}< \dfrac{1}{4.5}\)
\(\dfrac{1}{6^2}< \dfrac{1}{5.6}\)
......
\(\dfrac{1}{100^2}< \dfrac{1}{99.100}\)
\(\Rightarrow\dfrac{1}{5^2}+\dfrac{1}{6^2}+\dfrac{1}{7^2}+...+\dfrac{1}{100^2}< \dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{99.100}=\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{99}-\dfrac{1}{100}=\dfrac{1}{4}-\dfrac{1}{100}< \dfrac{1}{4}\)
6 tháng 9 2021
Ta có: \(\dfrac{1}{5^2}+\dfrac{1}{6^2}+\dfrac{1}{7^2}+...+\dfrac{1}{100^2}< \dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{99.100}\)
\(=\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(=\dfrac{1}{4}-\dfrac{1}{100}< \dfrac{1}{4}\)
TG
11 tháng 4 2022
1/4^2 + 1/5^2 +... + 1/100^2 < 1/3.4 + 1/4.5 +...+ 1/99.100
A=1/3 - 1/4 + 1/4 - 1/5 +...+ 1/99 - 1/100
=1/3 - 1/100 < 1/3
NT
26 tháng 12 2017
\(A=\dfrac{1}{5^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}\\ =\dfrac{1}{5.5}+\dfrac{1}{6.6}+...+\dfrac{1}{100.100}\\ < \dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{99.100}\\ =\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{99}-\dfrac{1}{100}\\ =\dfrac{1}{4}-\dfrac{1}{100}< \dfrac{1}{4}\)
\(A=\dfrac{1}{5^2}+\dfrac{1}{6^2}+....+\dfrac{1}{100^2}>\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{100.101}\\=\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{100}-\dfrac{1}{101}\\ =\dfrac{1}{5}-\dfrac{1}{101}\)
PT
2 tháng 3 2017
Giải:
Ta có:
\(\dfrac{1}{5^2}+\dfrac{1}{6^2}+\dfrac{1}{7^2}+...+\dfrac{1}{100^2}< \dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{99.100}\)
Đặt \(A=\dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{99.100}\)
\(B=\dfrac{1}{5^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}\)
\(A=\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(A=\dfrac{1}{4}-\dfrac{1}{100}\)
\(A=\dfrac{6}{25}\)
Mà \(\dfrac{1}{6}< \dfrac{6}{25}< \dfrac{1}{4}\)
Ta lại có \(A< \dfrac{6}{25}\)
Vậy \(\dfrac{1}{6}< \dfrac{1}{5^2}+\dfrac{1}{6^2}+\dfrac{1}{7^2}+...+\dfrac{1}{100^2}< \dfrac{1}{4}\)
16 tháng 4 2017
1/5^2< 1/4.5=1/4-1/5
1/6^2<1/5.6=1/5-1/6
..
1/99^2<1/98.99=1/98-1/99
1/100^2<1/99.100=1/99-1/100
Cộng vế theo vế, đơn giản:
=> 1/5^2+1/6^2+...+1/100^2< 1/4 -1/100<1/4
**
1/5^2> 1/5.6=1/5-1/6
1/6^2>1/6.7=1/6-1/7
..
1/99^2>1/99.100=1/99-1/100
1/100^2>1/100.101=1/100-1/101
Cộng vế theo vế, đơn giản:
=> 1/5^2+1/6^2+...+1/100^2>1/5 -1/101=96/505>1/6
Vậy:
1/6<1/5^2+1/6^2+...+1/100^2<1/4
21 tháng 9 2018
1/6<1/5^2+1/6^2+1/7^2+...+1/100^2<1/4
ta có:
(+)1/5^2+1/6^2+1/7^2+...+1/100^2<1/4.5+1/5.6+...+1/99.100
=1/4-1/5+1/5-...+1/99-1/100
=1/4-1/100<1/4
=>1/5^2+1/6^2+1/7^2+...+1/100^2<1/4
(+)1/5^2+1/6^2+1/7^2+...+1/100^2>1/5.6+...+1/99.100
=1/5-1/6+1/6-...+1/99-1/100
=1/5-1/100>1/6
=>1/5^2+1/6^2+1/7^2+...+1/100^2
NT
10 tháng 2 2018
\(\dfrac{1}{5^2}+\dfrac{1}{6^2}+\dfrac{1}{7^2}+....+\dfrac{1}{100^2}\\ >\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{100.101}\\ =\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{100}-\dfrac{1}{101}\\ =\dfrac{1}{5}-\dfrac{1}{101}\\ =\dfrac{96}{505}\\ >\dfrac{1}{6}\)
\(\dfrac{1}{5^2}+...+\dfrac{1}{100^2}\\ < \dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+....+\dfrac{1}{99.100}\\ =\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{99}-\dfrac{1}{100}\\ =\dfrac{1}{4}-\dfrac{1}{100}< \dfrac{1}{4}\)
23 tháng 2 2021
\(\dfrac{1}{4^2}+\dfrac{1}{5^2}+...+\dfrac{1}{100^2}>\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+...+\dfrac{1}{100\cdot101}=\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{100}-\dfrac{1}{101}=\dfrac{1}{4}-\dfrac{1}{101}>\dfrac{1}{4}-\dfrac{1}{20}=\dfrac{1}{5}\left(1\right)\)
\(\dfrac{1}{4^2}+\dfrac{1}{5^2}+...+\dfrac{1}{100^2}< \dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+...+\dfrac{1}{99\cdot100}=\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{100}=\dfrac{1}{3}-\dfrac{1}{100}< \dfrac{1}{3}\left(2\right)\) Từ (1) và (2) \(\Rightarrow\dfrac{1}{5}< \dfrac{1}{4^2}+\dfrac{1}{5^2}+...+\dfrac{1}{99^2}+\dfrac{1}{100^2}< \dfrac{1}{3}\)
NN
1
NV
Nguyễn Việt Lâm
Giáo viên
25 tháng 7 2021
Đặt \(A=\dfrac{1}{5^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}\)
Ta có: \(\dfrac{1}{5^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}< \dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{99.100}\)
\(\Rightarrow A< \dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(\Rightarrow A< \dfrac{1}{4}-\dfrac{1}{100}< \dfrac{1}{4}\) (đpcm)