Toán Chứng minh c, 1/2^2 + 1/3^2 + 1/4^2 + ….. + 1/10^2 < 1 11/09/2021 By Eden Chứng minh c, 1/2^2 + 1/3^2 + 1/4^2 + ….. + 1/10^2 < 1
Ta có: $\dfrac{1}{2^2}=\dfrac{1}{2.2}<\dfrac{1}{1.2} =\dfrac{2-1}{1.2}=1 -\dfrac12$ $\dfrac{1}{3^2}=\dfrac{1}{3.3}<\dfrac{1}{2.3}=\dfrac{3-2}{2.3}=\dfrac12-\dfrac13$ $\cdots$ $\dfrac{1}{10^2}=\dfrac{1}{10.10}<\dfrac{1}{9.10}=\dfrac{10 -9}{9.10}=\dfrac{1}{9}-\dfrac{1}{10}$ Cộng vế theo vế ta được: $\dfrac{1}{2^2}+\dfrac{1}{3^2}+\cdots +\dfrac{1}{10^2}<1-\dfrac{1}{2}+\dfrac12-\dfrac{1}{3}+\cdots+\dfrac19 -\dfrac{1}{10}$ $\Leftrightarrow \dfrac{1}{2^2}+\dfrac{1}{3^2}+\cdots +\dfrac{1}{10^2}<1-\dfrac{1}{10} < 1$ Vậy $\dfrac{1}{2^2}+\dfrac{1}{3^2}+\cdots +\dfrac{1}{10^2}<1$ Trả lời
Đáp án: `1/2^2 + 1/3^2 + 1/4^2 + … + 1/10^2 < 1` Giải thích các bước giải: Ta thấy : `1/2^2 < 1 – 1/2` `1/2^3 < 1/2-1/3` `1/4^2 < 1/3 – 1/4` ` . . . ` `1/10^2 < 1/9 – 1/10` `⇒1/2^2 + 1/2^3 + 1/4^2 + … + 1/10^2 < 1 – 1/2 + 1/2 – 1/3 + … + 1/9 – 1/10` `⇒1/2^2 + 1/2^3 + 1/4^2 + … + 1/10^2 < 1 – 1/10 < 1` Trả lời
Ta có:
$\dfrac{1}{2^2}=\dfrac{1}{2.2}<\dfrac{1}{1.2} =\dfrac{2-1}{1.2}=1 -\dfrac12$
$\dfrac{1}{3^2}=\dfrac{1}{3.3}<\dfrac{1}{2.3}=\dfrac{3-2}{2.3}=\dfrac12-\dfrac13$
$\cdots$
$\dfrac{1}{10^2}=\dfrac{1}{10.10}<\dfrac{1}{9.10}=\dfrac{10 -9}{9.10}=\dfrac{1}{9}-\dfrac{1}{10}$
Cộng vế theo vế ta được:
$\dfrac{1}{2^2}+\dfrac{1}{3^2}+\cdots +\dfrac{1}{10^2}<1-\dfrac{1}{2}+\dfrac12-\dfrac{1}{3}+\cdots+\dfrac19 -\dfrac{1}{10}$
$\Leftrightarrow \dfrac{1}{2^2}+\dfrac{1}{3^2}+\cdots +\dfrac{1}{10^2}<1-\dfrac{1}{10} < 1$
Vậy $\dfrac{1}{2^2}+\dfrac{1}{3^2}+\cdots +\dfrac{1}{10^2}<1$
Đáp án:
`1/2^2 + 1/3^2 + 1/4^2 + … + 1/10^2 < 1`
Giải thích các bước giải:
Ta thấy :
`1/2^2 < 1 – 1/2`
`1/2^3 < 1/2-1/3`
`1/4^2 < 1/3 – 1/4`
` . . . `
`1/10^2 < 1/9 – 1/10`
`⇒1/2^2 + 1/2^3 + 1/4^2 + … + 1/10^2 < 1 – 1/2 + 1/2 – 1/3 + … + 1/9 – 1/10`
`⇒1/2^2 + 1/2^3 + 1/4^2 + … + 1/10^2 < 1 – 1/10 < 1`