CM : \(\frac{1}{6}<\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+......+\frac{1}{100^2}<\frac{1}{4}\)
Những câu hỏi liên quan
X=\(\frac{1}{2}.\frac{3}{4}.\frac{5}{6}.....\frac{37}{38}\). cm\(\frac{1}{9}< x< \frac{1}{6}\)
CM
\(\frac{1}{6}<\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+....+\frac{1}{100^2}\)
CM : \(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+...+\frac{1}{16}+\frac{1}{17}< 2\)
\(A=\left(\frac{1}{5}+\frac{1}{6}+...+\frac{1}{9}\right)+\left(\frac{1}{10}+\frac{1}{11}+...+\frac{1}{17}\right)\)
\(\Rightarrow A< \left(\frac{1}{5}+\frac{1}{5}+...+\frac{1}{5}\right)+\left(\frac{1}{10}+\frac{1}{10}+...+\frac{1}{10}\right)\)
( 5 số hạng 1/5 ; 8 số hạng 1/10 )
\(\)\(\Rightarrow A< 5\cdot\frac{1}{5}+8\cdot\frac{1}{10}\)
\(\Rightarrow A< 2\)
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\(CM:\frac{1}{51}+\frac{1}{52}+....+\frac{1}{100}< \frac{5}{6}\)
\(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}< \frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}=50.\frac{1}{100}=\frac{1}{2}< \frac{5}{6}\)
Vậy \(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}< \frac{5}{6}\)
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CM: \(\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{100^2}< \frac{1}{2}\)
CM\(\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{100^2}< \frac{1}{2}\)
\(\frac{1}{2^2}< \frac{1}{1.2}\)
\(\frac{1}{4^2}< \frac{1}{3.4}\)
\(...\)
\(\frac{1}{100^2}< \frac{1}{99.100}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{1}{1.2}+...+\frac{1}{99.100}\)
Mình chỉ làm được đến đây thôi. Sorry nha. À mà bạn thử vào trang này xem https://vn.answers.yahoo.com/question/index?qid=20121102064330AAkYsXP
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CM: \(\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{100^2}< \frac{1}{4}\)
1/5^2 + 1/6^2 + 1/7^2 + ... + 1/100^2
< 1/4×5 + 1/5×6 + 1/6×7 + ... + 1/99×100
< 1/4 - 1/5 + 1/5 - 1/6 + 1/6 - 1/7 + ... + 1/99 - 1/100
< 1/4 - 1/100 < 1/4 ( đpcm)
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\(Cm:\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{2n-1}{2n}< \frac{2}{\sqrt{2n+1}}\)
CM \(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+....+\frac{1}{199}-\frac{1}{200}=\frac{1}{101}+\frac{1}{102}+....+\frac{1}{200}\)
\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{199}-\frac{1}{200}\)
\(=\left(1+\frac{1}{3}+...+\frac{1}{199}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{200}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{199}+\frac{1}{200}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{200}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{199}+\frac{1}{200}\right)-\left(1+\frac{1}{2}+...+\frac{1}{100}\right)\)
\(=\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\)
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Ta có :
\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{199}-\frac{1}{200}\)
\(=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{199}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{200}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{199}+\frac{1}{200}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{200}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{199}+\frac{1}{200}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)\)
\(=\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\left(đpcm\right)\)
Chúc bạn học tốt !!!
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