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DN
17 tháng 2 2018
\(S=\frac{1}{5}+\frac{2}{5^2}+\frac{3}{5^3}+...+\frac{2014}{5^{2014}}\)
\(5S=1+\frac{2}{5}+\frac{3}{5^2}+...+\frac{2014}{5^{2013}}\)
\(\Rightarrow5S-S=1+\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{2013}}-\frac{2014}{5^{2014}}\)
\(S=\frac{1+\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{2013}}-\frac{2014}{5^{2014}}}{4}\)
Xét \(A=\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{2013}}\)
\(5A=1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{2012}}\)
\(5A-A=1-\frac{1}{5^{2013}}\Leftrightarrow A=\frac{1-\frac{1}{5^{2013}}}{4}=\frac{1}{4}-\frac{1}{4.5^{2013}}\)
\(\Rightarrow S=\frac{1+\frac{1}{4}-\left(\frac{1}{4.5^{2013}}+\frac{2014}{5^{2014}}\right)}{4}=\frac{5}{16}-\frac{\frac{1}{4.5^{2013}}+\frac{2014}{5^{2014}}}{4}< \frac{1}{3}\)
31 tháng 3 2018
Ta có :
\(S=\frac{1}{5^2}+\frac{1}{5^4}+\frac{1}{5^6}+\frac{...1}{5^{2018}}\)
\(25S=1+\frac{1}{5^2}+\frac{1}{5^4}+...+\frac{1}{5^{2016}}\)
\(25S-S=\left(1+\frac{1}{5^2}+\frac{1}{5^4}+...+\frac{1}{5^{2016}}\right)-\left(\frac{1}{5^2}+\frac{1}{5^4}+\frac{1}{5^6}+...+\frac{1}{5^{2018}}\right)\)
\(24S=1-\frac{1}{5^{2018}}\)
\(S=\frac{1-\frac{1}{5^{2018}}}{24}\)
\(S=\frac{\frac{5^{2018}-1}{5^{2018}}}{24}< \frac{1}{24}\)
Vậy \(S< \frac{1}{24}\)
Chúc bạn học tốt ~
13 tháng 3 2018
\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2013}-\frac{1}{2014}\)
\(=\left(1+\frac{1}{3}+...+\frac{1}{2013}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2014}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2013}+\frac{1}{2014}-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2014}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2013}+\frac{1}{2014}-1-\frac{1}{2}-...-\frac{1}{1007}\)
\(=\frac{1}{1008}+\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2014}\) (đpcm)
6 tháng 3 2020
\(1\frac{1}{30}:\left(24\frac{1}{6}-24\frac{1}{5}\right)-\frac{1\frac{1}{2}-\frac{3}{4}}{4x-\frac{1}{2}}=\left(-1\frac{1}{15}\right):\left(8\frac{1}{5}-8\frac{1}{3}\right)\)
\(\Rightarrow\frac{31}{30}:\left(\frac{145}{6}-\frac{121}{5}\right)-\frac{\frac{3}{2}-\frac{3}{4}}{4x-\frac{1}{2}}=\left(\frac{-16}{15}\right):\left(\frac{41}{5}-\frac{25}{3}\right)\)
\(\Rightarrow\frac{31}{30}:\left(\frac{725}{30}-\frac{726}{30}\right)-\frac{\frac{6}{4}-\frac{3}{4}}{4x-\frac{1}{2}}=\left(\frac{-16}{15}\right):\left(\frac{123}{15}-\frac{125}{15}\right)\)
\(\Rightarrow\frac{31}{30}:\frac{-1}{30}-\frac{\frac{3}{4}}{4x-\frac{1}{2}}=\frac{-16}{15}:\frac{-2}{15}\)
\(\Rightarrow\frac{31}{30}.\frac{30}{-1}-\frac{3}{4}:\left(4x-\frac{1}{2}\right)=\frac{-16}{15}.\frac{15}{-2}\)
\(\Rightarrow\left(-31\right)-\frac{3}{4}:\left(4x-\frac{1}{2}\right)=8\)
\(\Rightarrow\frac{3}{4}:\left(4x-\frac{1}{2}\right)=\left(-31\right)-8\)
\(\Rightarrow\frac{3}{4}:\left(4x-\frac{1}{2}\right)=\left(-39\right)\)
\(\Rightarrow4x-\frac{1}{2}=\frac{3}{4}:\left(-39\right)\)
\(\Rightarrow4x-\frac{1}{2}=\frac{3}{4}.\frac{-1}{39}\)
\(\Rightarrow4x-\frac{1}{2}=\frac{-1}{52}\)
\(\Rightarrow4x=\frac{-1}{52}+\frac{1}{2}\)
\(\Rightarrow4x=\frac{-1}{52}+\frac{26}{52}\)
\(\Rightarrow4x=\frac{25}{52}\)
\(\Rightarrow x=\frac{25}{52}:4\)
\(\Rightarrow x=\frac{25}{52}.\frac{1}{4}\)
\(\Rightarrow x=\frac{25}{208}\)
Vậy \(x=\frac{25}{208}\)
Chúc bn học tốt
\(S=\frac{1}{5^2}+\frac{1}{5^4}+...+\frac{1}{5^{2014}}\)
=> \(5^2S=1+\frac{1}{5^2}+...+\frac{1}{5^{2012}}\)
=> \(25S-S=\left(1+\frac{1}{5^2}+...+\frac{1}{5^{2012}}\right)-\left(\frac{1}{5^2}+\frac{1}{5^4}+...+\frac{1}{5^{2014}}\right)\)
=> \(24S=1-\frac{1}{5^{2014}}\)
=> \(S=\left(1-\frac{1}{5^{2014}}\right):24\)
=> \(S=\frac{1}{24}-\frac{1}{24.5^{2014}}< \frac{1}{24}\)
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