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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 \(A=\dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{99.100}\)

\(B=\dfrac{1}{5^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}\)

\(A=\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{100}\)

\(A=\dfrac{1}{4}-\dfrac{1}{100}\)

\(A=\dfrac{6}{25}\)

\(\dfrac{1}{6}< \dfrac{6}{25}< \dfrac{1}{4}\)

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

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: \(\dfrac{1}{5^2}+\dfrac{1}{6^2}+\dfrac{1}{7^2}+...+\dfrac{1}{100^2}< \dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{99.100}\)

    \(=\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{100}\)

    \(=\dfrac{1}{4}-\dfrac{1}{100}< \dfrac{1}{4}\)

11 tháng 4 2022

giúp mk với ;-;"

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

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) \(\Rightarrow\dfrac{1}{5}< \dfrac{1}{4^2}+\dfrac{1}{5^2}+...+\dfrac{1}{99^2}+\dfrac{1}{100^2}< \dfrac{1}{3}\)

NV
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)

A=1/3^2+1/4^2+1/5^2+1/6^2+...+1/100^2<1/2-1/3+1/3-1/4+...+1/99-1/100

=>A<1/2-1/100<1/2

NA
Ngoc Anh Thai
Giáo viên
11 tháng 4 2021

a)

\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{30^2}\\ < \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{29.30}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{29}-\dfrac{1}{30}\\ =1-\dfrac{1}{30}=\dfrac{29}{30}< 1\left(dpcm\right)\)

b)

 \(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}+...+\dfrac{1}{99}+\dfrac{1}{100}=\dfrac{1}{10}+\left(\dfrac{1}{11}+\dfrac{1}{12}+...+\dfrac{1}{99}+\dfrac{1}{100}\right)\\ >\dfrac{1}{10}+\dfrac{1}{100}+\dfrac{1}{100}+...+\dfrac{1}{100}=\dfrac{1}{10}+\dfrac{90}{100}\\ =\dfrac{110}{100}>1\left(đpcm\right).\)

NA
Ngoc Anh Thai
Giáo viên
11 tháng 4 2021

c)

\(\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+...+\dfrac{1}{17}\\ =\left(\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{9}\right)+\left(\dfrac{1}{10}+\dfrac{1}{11}+...+\dfrac{1}{17}\right)\\ < \dfrac{1}{5}.5+\dfrac{1}{8}.8=1+1=2\left(đpcm\right)\)

d) tương tự câu 1

6 tháng 3 2021

\(A=\dfrac{1}{5^2}+\dfrac{1}{6^2}+\dfrac{1}{7^2}+...+\dfrac{1}{2019^2}>\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}+\dfrac{1}{7\cdot8}+...+\dfrac{1}{2019\cdot2020}=\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+...+\dfrac{1}{2019}-\dfrac{1}{2020}=\dfrac{1}{5}-\dfrac{1}{2020}=\dfrac{404-1}{2020}=\dfrac{403}{2020}>\dfrac{40}{2020}=\dfrac{20}{101}\left(1\right)\) \(A=\dfrac{1}{5^2}+\dfrac{1}{6^2}+\dfrac{1}{7^2}+...+\dfrac{1}{2019^2}< \dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}+...+\dfrac{1}{2018\cdot2019}=\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{2018}-\dfrac{1}{2019}=\dfrac{1}{4}-\dfrac{1}{2019}=\dfrac{2019-4}{4\cdot2019}=\dfrac{2015}{4\cdot2019}< \dfrac{2019}{4\cdot2019}=\dfrac{1}{4}\left(2\right)\) Từ (1) và (2) \(\Rightarrow\dfrac{20}{101}< A< \dfrac{1}{4}\)