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\(A=1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\)
\(A=1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{99\cdot100}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(A=1-\frac{1}{100}\)
\(A=\frac{99}{100}< 2\left(đpcm\right)\)
\(\frac{3}{2^2}\cdot\frac{8}{3^2}\cdot\frac{15}{4^2}\cdot.....\cdot\frac{899}{30^2}\)
\(=\frac{1\cdot3}{2\cdot2}\cdot\frac{2\cdot4}{3\cdot3}\cdot\frac{3\cdot5}{4\cdot4}\cdot.....\cdot\frac{29\cdot31}{30\cdot30}\)
\(=\frac{1}{2}\cdot\frac{3}{2}\cdot\frac{2}{3}\cdot\frac{4}{3}\cdot\frac{3}{4}\cdot\frac{5}{4}\cdot....\cdot\frac{29}{30}\cdot\frac{31}{30}\)
\(=\left(\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot....\cdot\frac{29}{30}\right)\cdot\left(\frac{3}{2}\cdot\frac{4}{3}\cdot\frac{5}{4}\cdot....\cdot\frac{31}{30}\right)\)
\(=\frac{1}{30}\cdot\frac{31}{2}\)
\(=\frac{31}{60}\)
b, \(A=\frac{3}{10}+\frac{3}{11}+\frac{3}{12}+\frac{3}{13}+\frac{3}{14}\)
Ta có:
\(\frac{3}{15}< \frac{3}{10}=\frac{3}{10}\)
\(\frac{3}{15}< \frac{3}{11}< \frac{3}{10}\)
\(\frac{3}{15}< \frac{3}{12}< \frac{3}{10}\)
\(\frac{3}{15}< \frac{3}{13}< \frac{3}{10}\)
\(\frac{3}{15}< \frac{3}{14}< \frac{3}{10}\)
\(\Rightarrow\frac{3}{15}+\frac{3}{15}+\frac{3}{15}+\frac{3}{15}+\frac{3}{15}< \frac{3}{10}+\frac{3}{11}+\frac{3}{12}+\frac{3}{13}+\frac{3}{14}< \frac{3}{10}+\frac{3}{10}+\frac{3}{10}+\frac{3}{10}+\frac{3}{10}\)
\(\Rightarrow\frac{3\cdot5}{15}< A< \frac{3\cdot5}{10}\)
\(\Rightarrow1< A< \frac{15}{10}=\frac{3}{2}\)
Mà \(\frac{3}{2}< 2\)
\(\Rightarrow1< A< 2\)
c ,Ta có
\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{49}+\frac{1}{50}-2\cdot\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{25}\right)+\left(\frac{1}{26}+\frac{1}{27}+...+\frac{1}{50}\right)-\left(1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{25}\right)\)
\(=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{49}+\frac{1}{50}\)
a) Ta có: \(\frac{1}{2^2}< \frac{1}{1.2}\) ; \(\frac{1}{3^2}< \frac{1}{2.3}\) ; \(\frac{1}{4^2}< \frac{1}{3.4}\) ; ... ; \(\frac{1}{2010^2}< \frac{1}{2009.2010}\)
=> \(Vt< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2009.2010}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2009}-\frac{1}{2010}\)
\(=1-\frac{1}{2010}< 1\)
\(...\Leftrightarrow-\frac{1}{10}< x< \frac{3}{5}\)
\(-\frac{1}{10}< x\Rightarrow-\frac{1}{10}< \frac{10x}{10}\Rightarrow10x>1\Rightarrow x>\frac{1}{10}\) (*)
\(x< \frac{3}{5}\Rightarrow\frac{5x}{5}< \frac{3}{5}\Rightarrow5x< 3\Rightarrow x< \frac{3}{5}\)
Vậy \(\frac{1}{10}< x< \frac{3}{5}\)
\(\frac{1}{4}< \frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{50.51}< \frac{1}{2}\)
\(\Rightarrow\frac{1}{4}< \frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{50}-\frac{1}{51}< \frac{1}{2}\)
\(\Rightarrow\frac{1}{4}< \frac{1}{3}-\frac{1}{50}< \frac{1}{2}\)
\(\Rightarrow0,25< 0,3137...< 0,5\) ( Đpcm )
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