cho F=1/2^2+1/3^2+1/4^2+…+1/100^2 chứng minh rằng 99/202 10/08/2021 Bởi Ruby cho F=1/2^2+1/3^2+1/4^2+…+1/100^2 chứng minh rằng 99/202 { "@context": "https://schema.org", "@type": "QAPage", "mainEntity": { "@type": "Question", "name": " cho F=1/2^2+1/3^2+1/4^2+...+1/100^2 chứng minh rằng 99/202
$\dfrac{1}{2^2}<\dfrac{1}{1.2}$ ………. $\dfrac{1}{100^2}<\dfrac{1}{99.100}$ $⇒F<\dfrac{1}{1.2}+\dfrac{1}{2.3}+…+\dfrac{1}{99.100}$ $⇒F<\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+…+\dfrac{1}{99}-\dfrac{1}{100}$ $⇒F<1-\dfrac{1}{100}=\dfrac{99}{100}$ $⇒F<\dfrac{99}{100}$ (1) $\dfrac{1}{2^2}>\dfrac{1}{2.3}$ ……… $\dfrac{1}{100^2}>\dfrac{1}{100.101}$ $⇒F>\dfrac{1}{2.3}+\dfrac{1}{3.4}+…+\dfrac{1}{100.101}$ $⇒F>\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+…+\dfrac{1}{100}-\dfrac{1}{101}$ $⇒F>\dfrac{1}{2}-\dfrac{1}{101}=\dfrac{99}{202}$ $⇒F>\dfrac{99}{202}$ (2) Từ (1), (2) $⇒\dfrac{99}{202}<F<\dfrac{99}{100}$ Bình luận
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$\dfrac{1}{2^2}<\dfrac{1}{1.2}$
……….
$\dfrac{1}{100^2}<\dfrac{1}{99.100}$
$⇒F<\dfrac{1}{1.2}+\dfrac{1}{2.3}+…+\dfrac{1}{99.100}$
$⇒F<\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+…+\dfrac{1}{99}-\dfrac{1}{100}$
$⇒F<1-\dfrac{1}{100}=\dfrac{99}{100}$
$⇒F<\dfrac{99}{100}$ (1)
$\dfrac{1}{2^2}>\dfrac{1}{2.3}$
………
$\dfrac{1}{100^2}>\dfrac{1}{100.101}$
$⇒F>\dfrac{1}{2.3}+\dfrac{1}{3.4}+…+\dfrac{1}{100.101}$
$⇒F>\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+…+\dfrac{1}{100}-\dfrac{1}{101}$
$⇒F>\dfrac{1}{2}-\dfrac{1}{101}=\dfrac{99}{202}$
$⇒F>\dfrac{99}{202}$ (2)
Từ (1), (2) $⇒\dfrac{99}{202}<F<\dfrac{99}{100}$