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3 tháng 5

\(\dfrac{1}{3^2}+\dfrac{1}{4^2}+\dfrac{1}{5^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}\\ =\dfrac{1}{3.3}+\dfrac{1}{4.4}+\dfrac{1}{5.5}+\dfrac{1}{6.6}+...+\dfrac{1}{100.100}\)

Ta có: \(\dfrac{1}{3.3}+\dfrac{1}{4.4}+\dfrac{1}{5.5}+\dfrac{1}{6.6}+...+\dfrac{1}{100.100}< \dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{99.100}\\ \dfrac{1}{3.3}+\dfrac{1}{4.4}+\dfrac{1}{5.5}+\dfrac{1}{6.6}+...+\dfrac{1}{100.100}< \dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{99}-\dfrac{1}{100}\\ \dfrac{1}{3.3}+\dfrac{1}{4.4}+\dfrac{1}{5.5}+\dfrac{1}{6.6}+...+\dfrac{1}{100.100}< \dfrac{1}{2}-\dfrac{1}{100}\\ \dfrac{1}{3.3}+\dfrac{1}{4.4}+\dfrac{1}{5.5}+\dfrac{1}{6.6}+...+\dfrac{1}{100.100}< \dfrac{49}{100}< \dfrac{50}{100}=\dfrac{1}{2}\)

Hay \(\dfrac{1}{3.3}+\dfrac{1}{4.4}+\dfrac{1}{5.5}+\dfrac{1}{6.6}+...+\dfrac{1}{100.100}< \dfrac{1}{2}\) 

Vì \(\dfrac{1}{3.3}+\dfrac{1}{4.4}+\dfrac{1}{5.5}+\dfrac{1}{6.6}+...+\dfrac{1}{100.100}=\dfrac{1}{3^2}+\dfrac{1}{4^2}+\dfrac{1}{5^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}\)

Vậy biểu thức \(\dfrac{1}{3^2}+\dfrac{1}{4^2}+\dfrac{1}{5^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}< \dfrac{1}{2}\)

3 tháng 5

1/3² + 1/4² + 1/5² + 1/6² + ... + 1/100²

< 1/(2.3) + 1/(3.4) + 1/(4.5) + 1/(5.6) + ... + 1/(99.100)

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

= 1/2 - 1/100 < 1/2

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

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

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

29 tháng 5 2018

Ta có :

\(\dfrac{1}{3^2}< \dfrac{1}{2.3}\)

\(\dfrac{1}{4^2}< \dfrac{1}{3.4}\)

...................

\(\dfrac{1}{100^2}< \dfrac{1}{99.100}\)

\(\Leftrightarrow\dfrac{1}{3^2}+\dfrac{1}{4^2}+..........+\dfrac{1}{100^2}< \dfrac{1}{2.3}+\dfrac{1}{3.4}+.....+\dfrac{1}{99.100}\)

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

\(\Leftrightarrow\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}< \dfrac{1}{2}-\dfrac{1}{100}\)

\(\Leftrightarrow\dfrac{1}{3^2}+\dfrac{1}{4^2}+.......+\dfrac{1}{100^2}< \dfrac{49}{100}< \dfrac{1}{2}\)

29 tháng 5 2018

Tham khảo : https://olm.vn/hoi-dap/question/564353.html

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 5 2022

ơi

11 tháng 5 2022

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)