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\(M=\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{2013}}+\frac{1}{5^{2014}}\)
\(5M=1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{2013}}\)
\(\Rightarrow4M=1-\frac{1}{5^{2014}}< 1\)
\(\Rightarrow M< \frac{1}{4}< \frac{1}{3}\)
Đề sai, đề đúng phải là \(VT< \frac{1}{20}\)
Dễ dàng chứng minh đề sai, ta có:
\(\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{2006}}>\frac{1}{5^2}+\frac{1}{5^3}=\frac{6}{125}>\frac{1}{24}\)
Còn chứng minh \(VT< \frac{1}{20}\) thì như sau:
\(A=\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{2006}}\)
\(\Rightarrow5A=\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{2005}}\)
\(\Rightarrow5A-\frac{1}{5}+\frac{1}{5^{2006}}=\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{2006}}\)
\(\Rightarrow5A-\frac{1}{5}+\frac{1}{5^{2006}}=A\)
\(\Rightarrow4A=\frac{1}{5}-\frac{1}{5^{2006}}< \frac{1}{5}\)
\(\Rightarrow A< \frac{1}{20}\)
\(A=\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+...+\frac{1}{97\cdot99}-\frac{5}{4}\cdot\frac{13}{99}+\frac{5}{99}\cdot\frac{1}{4}\)
\(A=\frac{1}{2}\left(\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{97\cdot99}\right)-\frac{13}{4}\cdot\frac{5}{99}+\frac{5}{99}\cdot\frac{1}{4}\)
\(A=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)-\frac{5}{99}\cdot\left(\frac{13}{4}-\frac{1}{4}\right)\)
\(A=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{99}\right)-\frac{5}{99}\cdot3\)
\(A=\frac{1}{2}\cdot\frac{32}{99}-\frac{5}{33}\)
\(A=\frac{16}{99}-\frac{5}{33}=\frac{1}{99}\)
Ta có: \(B=\frac{1}{5^2}+\frac{1}{5^4}+\frac{1}{5^6}+...+\frac{1}{5^{2014}}\)
=> \(25B=1+\frac{1}{5^2}+\frac{1}{5^4}+...+\frac{1}{5^{2012}}\)
=> 25B-B=24B= \(1-\frac{1}{5^{2014}}\)
=> \(B=\frac{1-\frac{1}{5^{2014}}}{24}< \frac{1}{24}\)
=> đpcm