Chứng minh : S = \(\frac{1}{5}\)+ \(\frac{1}{13}\)+ \(\frac{1}{14}\)+ \(\frac{1}{15}\)+ \(\frac{1}{61}\)+\(\frac{1}{62}\)+\(\frac{1}{63}\)< \(\frac{1}{2}\)
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có : \(S=\frac{1}{5}+\left(\frac{1}{13}+\frac{1}{14}+\frac{1}{15}\right)+\left(\frac{1}{61}+\frac{1}{62}+\frac{1}{63}\right)
Ta có:
\(S=\frac{1}{5}+\frac{1}{13}+\frac{1}{14}+\frac{1}{15}+\frac{1}{61}+\frac{1}{62}+\frac{1}{63}\)
\(=\frac{1}{5}+\left(\frac{1}{13}+\frac{1}{14}+\frac{1}{15}\right)+\left(\frac{1}{61}+\frac{1}{62}+\frac{1}{63}\right)\)
Bài toán phụ 1:
Ta có:
1/13<1/12
1/14<1/12
1/15<1/12
=>1/13+1/14+1/15<1/12x3=1/4 (1)
Bài toán phụ 2:
Ta có:
1/61<1/60
1/62<1/60
1/63<1/60
=>1/61+1/62+1/63<1/60x3=1/20 (2)
Từ (1) và (2), ta có:
1/5+1/13+1/14+1/15+1/61+1/62+1/63<1/5+1/4+1/20
1/5+1/13+1/14+1/15+1/61+1/62+1/63<4/20+5/20+1/20
1/5+1/13+1/14+1/15+1/61+1/62+1/63<9/20<1/2
=>1/5+1/13+1/14+1/15+1/61+1/62+1/63<1/2
![](https://rs.olm.vn/images/avt/0.png?1311)
S=1/5+(1/13+1/14+1/15)+(1/61+1/62+1/63)
suy ra S<1/5+1/12.3+1/60.3
S<1/5+1/4+1/20
S<1/2
Dãy số này không viết theo qui luật nên bạn chỉ có cách là cộng tất cả các số hạng vào rồi xét nó bé hơn \(\frac{1}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
A = 1/5 + 1/13 + 1/14 + 1/15 + 1/60 + 1/61 + 1/62 + 1/63
Ta có : A = 1/5 + 1/13 + 1/14 + 1/15 + 1/60 + 1/61 + 1/62 + 1/63 < 1/5 + 1/12 + 1/12 + 1/12 + 1/60 + 1/60 + 1/60
= A < 1/5 + 1/4 + 1/20
= A < 1/2
Vậy A < 1/12
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có :
\(\frac{1}{13}< \frac{1}{12};\frac{1}{14}< \frac{1}{12};\frac{1}{15}< \frac{1}{12}\Rightarrow\frac{1}{13}+\frac{1}{14}+\frac{1}{15}< \frac{1}{12}+\frac{1}{12}+\frac{1}{12}=\frac{1}{4}\)
\(\frac{1}{61}< \frac{1}{60};\frac{1}{62}< \frac{1}{60};\frac{1}{63}< \frac{1}{60}\Rightarrow\frac{1}{61}+\frac{1}{62}+\frac{1}{63}< \frac{1}{60}+\frac{1}{60}+\frac{1}{60}=\frac{1}{20}\)
\(\Rightarrow D=\frac{1}{5}+\left(\frac{1}{13}+\frac{1}{14}+\frac{1}{15}\right)+\left(\frac{1}{61}+\frac{1}{62}+\frac{1}{63}\right)< \frac{1}{5}+\frac{1}{4}+\frac{1}{20}=\frac{1}{2}\)
Vậy \(D< \frac{1}{2}\)
\(D=\frac{1}{5}+\frac{1}{13}+\frac{1}{14}+\frac{1}{15}+\frac{1}{61}+\frac{1}{62}+\frac{1}{63}=\frac{1}{5}+\left(\frac{1}{13}+\frac{1}{14}+\frac{1}{15}\right)+\left(\frac{1}{61}+\frac{1}{62}+\frac{1}{63}\right)\)
Nhận xét: \(\frac{1}{13}+\frac{1}{14}+\frac{1}{15}< \frac{1}{12}+\frac{1}{12}+\frac{1}{12}=\frac{3}{12}=\frac{1}{4}\)
\(\frac{1}{61}+\frac{1}{62}+\frac{1}{63}< \frac{1}{60}+\frac{1}{60}+\frac{1}{60}=\frac{3}{60}=\frac{1}{20}\)
\(\Rightarrow D< \frac{1}{5}+\frac{1}{4}+\frac{1}{20}=\frac{1}{2}\)
Vậy D < 1/2
![](https://rs.olm.vn/images/avt/0.png?1311)
a,Ta có: \(\frac{3}{10}=\frac{3}{10};\frac{3}{11}< \frac{3}{10};\frac{3}{12}< \frac{3}{10};\frac{3}{13}< \frac{3}{10};\frac{3}{14}< \frac{3}{10}\)
\(\Rightarrow S< \frac{3}{10}+\frac{3}{10}+\frac{3}{10}+\frac{3}{10}+\frac{3}{10}=\frac{15}{10}=\frac{3}{2}=1,5\left(1\right)\)
Lại có: \(\frac{3}{10}>\frac{3}{15};\frac{3}{11}>\frac{3}{15};\frac{3}{12}>\frac{3}{15};\frac{3}{13}>\frac{3}{15};\frac{3}{14}>\frac{3}{15}\)
\(\Rightarrow S>\frac{3}{15}+\frac{3}{15}+\frac{3}{15}+\frac{3}{15}+\frac{3}{15}=\frac{15}{15}=1\left(2\right)\)
Từ (1) và (2) => 1 < S < 1,5
Vậy...
b, \(A=\frac{1}{61}+\frac{1}{62}+...+\frac{1}{100}\)
\(=\left(\frac{1}{61}+\frac{1}{62}+...+\frac{1}{80}\right)+\left(\frac{1}{81}+\frac{1}{82}+...+\frac{1}{100}\right)\)
Ta có: \(\frac{1}{61}>\frac{1}{80};\frac{1}{62}>\frac{1}{80};...;\frac{1}{80}=\frac{1}{80}\)
\(\Rightarrow\frac{1}{61}+\frac{1}{62}+...+\frac{1}{80}>\frac{1}{80}+\frac{1}{80}+...+\frac{1}{80}=\frac{20}{80}=\frac{1}{4}\left(1\right)\)
Lại có: \(\frac{1}{81}>\frac{1}{100};\frac{1}{82}>\frac{1}{100};...;\frac{1}{100}=\frac{1}{100}\)
\(\Rightarrow\frac{1}{81}+\frac{1}{82}+...+\frac{1}{100}>\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}=\frac{20}{100}=\frac{1}{5}\left(2\right)\)
Từ (1) và (2) => \(A>\frac{1}{4}+\frac{1}{5}=\frac{9}{20}\)
Vậy...