Câu hỏi của Cao Minh Quân

Question

CỨU TUI MN ƠI

CMR 1/6<1/(5^2)+1/(6^2)+1/(7^2)+...+1/(100^2)<1/4

Nguyễn Lê Phước Thịnh · Accepted Answer

Sửa đề: $\dfrac{1}{5}< \dfrac{1}{5^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}< \dfrac{1}{4}$Đặt $A=\dfrac{1}{5^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}$$\dfrac{1}{5}-\dfrac{1}{6}< \dfrac{1}{5\cdot6}< \dfrac{1}{5^2}< \dfrac{1}{4\cdot5}=\dfrac{1}{4}-\dfrac{1}{5}$$\dfrac{1}{6}-\dfrac{1}{7}< \dfrac{1}{6\cdot7}< \dfrac{1}{6^2}< \dfrac{1}{5\cdot6}=\dfrac{1}{5}-\dfrac{1}{6}$...$\dfrac{1}{100}-\dfrac{1}{101}=\dfrac{1}{100\cdot101}< \dfrac{1}{100^2}< \dfrac{1}{100\cdot99}=\dfrac{1}{99}-\dfrac{1}{100}$Do đó: $\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{100}-\dfrac{1}{101}< \dfrac{1}{5^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}< \dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{99}-\dfrac{1}{100}$=>$\dfrac{1}{5}-\dfrac{1}{101}< A< \dfrac{1}{4}-\dfrac{1}{100}$=>$\dfrac{1}{5}< A< \dfrac{1}{4}$Câu trả lời: Sửa đề: $\dfrac{1}{5}< \dfrac{1}{5^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}< \dfrac{1}{4}$Đặt $A=\dfrac{1}{5^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}$$\dfrac{1}{5}-\dfrac{1}{6}< \dfrac{1}{5\cdot6}< \dfrac{1}{5^2}< \dfrac{1}{4\cdot5}=\dfrac{1}{4}-\dfrac{1}{5}$$\dfrac{1}{6}-\dfrac{1}{7}< \dfrac{1}{6\cdot7}< \dfrac{1}{6^2}< \dfrac{1}{5\cdot6}=\dfrac{1}{5}-\dfrac{1}{6}$...$\dfrac{1}{100}-\dfrac{1}{101}=\dfrac{1}{100\cdot101}< \dfrac{1}{100^2}< \dfrac{1}{100\cdot99}=\dfrac{1}{99}-\dfrac{1}{100}$Do đó: $\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{100}-\dfrac{1}{101}< \dfrac{1}{5^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}< \dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{99}-\dfrac{1}{100}$=>$\dfrac{1}{5}-\dfrac{1}{101}< A< \dfrac{1}{4}-\dfrac{1}{100}$=>$\dfrac{1}{5}< A< \dfrac{1}{4}$

Nguyễn Thị Thương Hoài · Answer

A = $\dfrac{1}{5^2}$ + $\dfrac{1}{6^2}$ + $\dfrac{1}{7^2}$ + ... + $\dfrac{1}{100^2}$ $\dfrac{1}{5.6}$ < $\dfrac{1}{5^2}$ < $\dfrac{1}{4.5}$ $\dfrac{1}{6.7}$ < $\dfrac{1}{6^2}$ < $\dfrac{1}{5.6}$ $\dfrac{1}{7.8}$ < $\dfrac{1}{7^2}$ < $\dfrac{1}{6.7}$ ...................... $\dfrac{1}{100.101}$ < $\dfrac{1}{100^2}$ < $\dfrac{1}{99.100}$ Cộng vế với vế ta có: $\dfrac{1}{5.6}$ + $\dfrac{1}{6.7}$ + ... + $\dfrac{1}{100.101}$< $\dfrac{1}{5^2}$+$\dfrac{1}{6^2}$+...+$\dfrac{1}{100^2}$<$\dfrac{1}{4.5}$+$\dfrac{1}{5.6}$+...+$\dfrac{1}{99.100}$ $\dfrac{1}{5}$-$\dfrac{1}{6}$+$\dfrac{1}{6}$-$\dfrac{1}{7}$+$\dfrac{1}{100}$-$\dfrac{1}{101}$ < $\dfrac{1}{5^2}$+$\dfrac{1}{6^2}$+...+$\dfrac{1}{100^2}$< $\dfrac{1}{4}$-$\dfrac{1}{5}$+$\dfrac{1}{5}$-$\dfrac{1}{6}$+...+$\dfrac{1}{99}$-$\dfrac{1}{100}$ $\dfrac{1}{5}$ - $\dfrac{1}{101}$ < $\dfrac{1}{5^2}$ + $\dfrac{1}{6^2}$+...+$\dfrac{1}{100^2}$ < $\dfrac{1}{4}$ - $\dfrac{1}{100}$ $\dfrac{6}{30}$ - $\dfrac{1}{101}$ < $\dfrac{1}{5^2}$ + $\dfrac{1}{6^2}$+ .... + $\dfrac{1}{100^2}$ < $\dfrac{1}{4}$ - $\dfrac{1}{100}$ < $\dfrac{1}{4}$ $\dfrac{5}{30}$ +( $\dfrac{1}{30}$ - $\dfrac{1}{101}$) < $\dfrac{1}{5^2}$ + $\dfrac{1}{6^2}$ + ... + $\dfrac{1}{100^2}$ < $\dfrac{1}{4}$ $\dfrac{1}{6}$ + ($\dfrac{1}{30}$ - $\dfrac{1}{101}$) < $\dfrac{1}{5^2}$ + $\dfrac{1}{6^2}$ + ... + $\dfrac{1}{100^2}$ < $\dfrac{1}{4}$ Vì $\dfrac{1}{30}$ > $\dfrac{1}{101}$ ⇒ $\dfrac{1}{30}$ - $\dfrac{1}{101}$ > 0 ⇒ $\dfrac{1}{6}$ + ($\dfrac{1}{30}$ - $\dfrac{1}{101}$) > $\dfrac{1}{6}$ Vậy $\dfrac{1}{6}$ < $\dfrac{1}{5^2}$ + $\dfrac{1}{6^2}$ + ... + $\dfrac{1}{100^2}$ < $\dfrac{1}{4}$ (đpcm)Câu trả lời: A = $\dfrac{1}{5^2}$ + $\dfrac{1}{6^2}$ + $\dfrac{1}{7^2}$ + ... + $\dfrac{1}{100^2}$ $\dfrac{1}{5.6}$ < $\dfrac{1}{5^2}$ < $\dfrac{1}{4.5}$ $\dfrac{1}{6.7}$ < $\dfrac{1}{6^2}$ < $\dfrac{1}{5.6}$ $\dfrac{1}{7.8}$ < $\dfrac{1}{7^2}$ < $\dfrac{1}{6.7}$ ...................... $\dfrac{1}{100.101}$ < $\dfrac{1}{100^2}$ < $\dfrac{1}{99.100}$ Cộng vế với vế ta có: $\dfrac{1}{5.6}$ + $\dfrac{1}{6.7}$ + ... + $\dfrac{1}{100.101}$< $\dfrac{1}{5^2}$+$\dfrac{1}{6^2}$+...+$\dfrac{1}{100^2}$<$\dfrac{1}{4.5}$+$\dfrac{1}{5.6}$+...+$\dfrac{1}{99.100}$ $\dfrac{1}{5}$-$\dfrac{1}{6}$+$\dfrac{1}{6}$-$\dfrac{1}{7}$+$\dfrac{1}{100}$-$\dfrac{1}{101}$ < $\dfrac{1}{5^2}$+$\dfrac{1}{6^2}$+...+$\dfrac{1}{100^2}$< $\dfrac{1}{4}$-$\dfrac{1}{5}$+$\dfrac{1}{5}$-$\dfrac{1}{6}$+...+$\dfrac{1}{99}$-$\dfrac{1}{100}$ $\dfrac{1}{5}$ - $\dfrac{1}{101}$ < $\dfrac{1}{5^2}$ + $\dfrac{1}{6^2}$+...+$\dfrac{1}{100^2}$ < $\dfrac{1}{4}$ - $\dfrac{1}{100}$ $\dfrac{6}{30}$ - $\dfrac{1}{101}$ < $\dfrac{1}{5^2}$ + $\dfrac{1}{6^2}$+ .... + $\dfrac{1}{100^2}$ < $\dfrac{1}{4}$ - $\dfrac{1}{100}$ < $\dfrac{1}{4}$ $\dfrac{5}{30}$ +( $\dfrac{1}{30}$ - $\dfrac{1}{101}$) < $\dfrac{1}{5^2}$ + $\dfrac{1}{6^2}$ + ... + $\dfrac{1}{100^2}$ < $\dfrac{1}{4}$ $\dfrac{1}{6}$ + ($\dfrac{1}{30}$ - $\dfrac{1}{101}$) < $\dfrac{1}{5^2}$ + $\dfrac{1}{6^2}$ + ... + $\dfrac{1}{100^2}$ < $\dfrac{1}{4}$ Vì $\dfrac{1}{30}$ > $\dfrac{1}{101}$ ⇒ $\dfrac{1}{30}$ - $\dfrac{1}{101}$ > 0 ⇒ $\dfrac{1}{6}$ + ($\dfrac{1}{30}$ - $\dfrac{1}{101}$) > $\dfrac{1}{6}$ Vậy $\dfrac{1}{6}$ < $\dfrac{1}{5^2}$ + $\dfrac{1}{6^2}$ + ... + $\dfrac{1}{100^2}$ < $\dfrac{1}{4}$ (đpcm)

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