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\(S=\frac{5}{20}+\frac{5}{21}+..........+\frac{5}{49}\)
\(=5\left(\frac{1}{20}+\frac{1}{21}+.......+\frac{1}{49}\right)\)
Mà \(\frac{1}{20}>\frac{1}{49};\frac{1}{21}>\frac{1}{49};.........;\frac{1}{49}=\frac{1}{49}\)
\(\Leftrightarrow5\left(\frac{1}{20}+\frac{1}{21}+.....+\frac{1}{49}\right)>5\left(\frac{1}{49}+\frac{1}{49}+.......+\frac{1}{49}\right)\)
\(\Leftrightarrow S>5.\frac{30}{49}\)
\(\Leftrightarrow S>3\frac{3}{49}\)
\(\Leftrightarrow S>3\left(1\right)\)
Lại có :
\(\frac{1}{20}=\frac{1}{20};\frac{1}{21}< \frac{1}{20};.......;\frac{1}{49}< \frac{1}{20}\)
\(\Leftrightarrow S=5\left(\frac{1}{20}+\frac{1}{21}+...+\frac{1}{49}\right)< 5\left(\frac{1}{20}+\frac{1}{20}+....+\frac{1}{20}\right)\)
\(\Leftrightarrow S< 5.\frac{30}{20}=7\frac{1}{2}\)
\(\Leftrightarrow S< 8\left(2\right)\)
Từ \(\left(1\right)+\left(2\right)\Leftrightarrow3< S< 8\)
\(S=5.\left(\frac{1}{20}+\frac{1}{21}+...+\frac{1}{49}\right)\)
Xét \(A=\frac{1}{20}+\frac{1}{21}+...+\frac{1}{49}\). Chứng minh 3/5 < A < 8/5
+ Có: \(\frac{1}{20}+\frac{1}{21}+...+\frac{1}{29}<\frac{1}{20}+\frac{1}{20}+...+\frac{1}{20}=\frac{1}{2}\)
\(\frac{1}{30}+\frac{1}{31}+...+\frac{1}{34}<\frac{1}{30}+\frac{1}{30}+...+\frac{1}{30}=\frac{15}{30}=\frac{1}{2}\)
\(\frac{1}{35}+\frac{1}{36}+...+\frac{1}{49}<\frac{1}{35}+\frac{1}{35}+...+\frac{1}{35}=\frac{15}{35}=\frac{3}{7}<\frac{3}{5}\)
Cộng từng vế => \(A<\frac{1}{2}+\frac{1}{2}+\frac{3}{5}=\frac{8}{5}\Rightarrow S<8\) (1)
+) Có :
\(\frac{1}{20}+\frac{1}{21}+\frac{1}{22}+\frac{1}{23}+\frac{1}{24}>\frac{1}{25}.5=\frac{1}{5}\)
\(\frac{1}{25}+\frac{1}{26}+...+\frac{1}{30}>\frac{1}{30}.6=\frac{1}{5}\)
\(\frac{1}{30}+...+\frac{1}{37}>\frac{1}{40}.8=\frac{1}{5}\)
=> \(\frac{1}{20}+...+\frac{1}{37}>\frac{1}{5}+\frac{1}{5}+\frac{1}{5}=\frac{3}{5}\)
=> \(A>\frac{1}{20}+...+\frac{1}{37}>\frac{3}{5}\Rightarrow S>3\) (2)
Từ (1)(2) => 3 < S < 8
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tham khảo nhé bn
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tham khảo nhé bn