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$\dfrac{1}{5}< \dfrac{1}{4^2}+\dfrac{1}{5^2}+...+\dfrac{1}{100^2}< \dfrac{1}{3}$

Nguyễn Thanh Hằng · Accepted Answer

Ta có : +) $\dfrac{1}{4^2}< \dfrac{1}{3.4}$ +) $\dfrac{1}{5^2}< \dfrac{1}{4.5}$ ....................... +) $\dfrac{1}{100^2}< \dfrac{1}{99.100}$ $\Leftrightarrow\dfrac{1}{4^2}+\dfrac{1}{5^2}+.....+\dfrac{1}{100^2}< \dfrac{1}{3.4}+\dfrac{1}{4.5}+....+\dfrac{1}{99.100}$ $\Leftrightarrow\dfrac{1}{4^2}+\dfrac{1}{5^2}+............+\dfrac{1}{100^2}< \dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+....+\dfrac{1}{99}-\dfrac{1}{100}$ $\Leftrightarrow\dfrac{1}{4^2}+\dfrac{1}{5^2}+......+\dfrac{1}{100}< \dfrac{1}{3}-\dfrac{1}{100}< \dfrac{1}{3}$ $\left(1\right)$ Lại có : +) $\dfrac{1}{4^2}>\dfrac{1}{4.5}$ +) $\dfrac{1}{5^2}>\dfrac{1}{5.6}$ ..................... +) $\dfrac{1}{100^2}>\dfrac{1}{100.101}$ $\Leftrightarrow\dfrac{1}{4^2}+\dfrac{1}{5^2}+........+\dfrac{1}{100^2}>\dfrac{1}{4.5}+\dfrac{1}{5.6}+.........+\dfrac{1}{100.101}$ $\Leftrightarrow\dfrac{1}{4^2}+\dfrac{1}{5^2}+........+\dfrac{1}{100^2}>\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+.....+\dfrac{1}{100}-\dfrac{1}{101}$ $\Leftrightarrow\dfrac{1}{4^2}+\dfrac{1}{5^2}+........+\dfrac{1}{100^2}>\dfrac{1}{4}-\dfrac{1}{101}>\dfrac{1}{4}>\dfrac{1}{5}\left(2\right)$ Từ $\left(1\right)+\left(2\right)\Leftrightarrowđpcm$Câu trả lời: Ta có : +) $\dfrac{1}{4^2}< \dfrac{1}{3.4}$ +) $\dfrac{1}{5^2}< \dfrac{1}{4.5}$ ....................... +) $\dfrac{1}{100^2}< \dfrac{1}{99.100}$ $\Leftrightarrow\dfrac{1}{4^2}+\dfrac{1}{5^2}+.....+\dfrac{1}{100^2}< \dfrac{1}{3.4}+\dfrac{1}{4.5}+....+\dfrac{1}{99.100}$ $\Leftrightarrow\dfrac{1}{4^2}+\dfrac{1}{5^2}+............+\dfrac{1}{100^2}< \dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+....+\dfrac{1}{99}-\dfrac{1}{100}$ $\Leftrightarrow\dfrac{1}{4^2}+\dfrac{1}{5^2}+......+\dfrac{1}{100}< \dfrac{1}{3}-\dfrac{1}{100}< \dfrac{1}{3}$ $\left(1\right)$ Lại có : +) $\dfrac{1}{4^2}>\dfrac{1}{4.5}$ +) $\dfrac{1}{5^2}>\dfrac{1}{5.6}$ ..................... +) $\dfrac{1}{100^2}>\dfrac{1}{100.101}$ $\Leftrightarrow\dfrac{1}{4^2}+\dfrac{1}{5^2}+........+\dfrac{1}{100^2}>\dfrac{1}{4.5}+\dfrac{1}{5.6}+.........+\dfrac{1}{100.101}$ $\Leftrightarrow\dfrac{1}{4^2}+\dfrac{1}{5^2}+........+\dfrac{1}{100^2}>\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+.....+\dfrac{1}{100}-\dfrac{1}{101}$ $\Leftrightarrow\dfrac{1}{4^2}+\dfrac{1}{5^2}+........+\dfrac{1}{100^2}>\dfrac{1}{4}-\dfrac{1}{101}>\dfrac{1}{4}>\dfrac{1}{5}\left(2\right)$ Từ $\left(1\right)+\left(2\right)\Leftrightarrowđpcm$

Nguyễn Việt Lâm · Answer

$\dfrac{1}{4^2}+\dfrac{1}{5^2}+...+\dfrac{1}{100^2}>\dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{100.101}$ $\Rightarrow\dfrac{1}{4^2}+\dfrac{1}{5^2}+...+\dfrac{1}{100^2}>\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{100}-\dfrac{1}{101}$ $\Rightarrow\dfrac{1}{4^2}+\dfrac{1}{5^2}+...+\dfrac{1}{100^2}>\dfrac{1}{4}-\dfrac{1}{101}>\dfrac{1}{4}-\dfrac{1}{20}=\dfrac{1}{5}$ Lại có $\dfrac{1}{4^2}+\dfrac{1}{5^2}+...+\dfrac{1}{100^2}< \dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{99.100}$ $\Rightarrow\dfrac{1}{4^2}+\dfrac{1}{5^2}+...+\dfrac{1}{100^2}< \dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{100}$ $\Rightarrow\dfrac{1}{4^2}+\dfrac{1}{5^2}+...+\dfrac{1}{100^2}< \dfrac{1}{3}-\dfrac{1}{100}< \dfrac{1}{3}$ Vậy $\dfrac{1}{5}< \dfrac{1}{4^2}+\dfrac{1}{5^2}+...+\dfrac{1}{100^2}< \dfrac{1}{3}$Câu trả lời: $\dfrac{1}{4^2}+\dfrac{1}{5^2}+...+\dfrac{1}{100^2}>\dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{100.101}$ $\Rightarrow\dfrac{1}{4^2}+\dfrac{1}{5^2}+...+\dfrac{1}{100^2}>\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{100}-\dfrac{1}{101}$ $\Rightarrow\dfrac{1}{4^2}+\dfrac{1}{5^2}+...+\dfrac{1}{100^2}>\dfrac{1}{4}-\dfrac{1}{101}>\dfrac{1}{4}-\dfrac{1}{20}=\dfrac{1}{5}$ Lại có $\dfrac{1}{4^2}+\dfrac{1}{5^2}+...+\dfrac{1}{100^2}< \dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{99.100}$ $\Rightarrow\dfrac{1}{4^2}+\dfrac{1}{5^2}+...+\dfrac{1}{100^2}< \dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{100}$ $\Rightarrow\dfrac{1}{4^2}+\dfrac{1}{5^2}+...+\dfrac{1}{100^2}< \dfrac{1}{3}-\dfrac{1}{100}< \dfrac{1}{3}$ Vậy $\dfrac{1}{5}< \dfrac{1}{4^2}+\dfrac{1}{5^2}+...+\dfrac{1}{100^2}< \dfrac{1}{3}$

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