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Ta có:
\(\frac{1}{5}=\frac{1}{5}\)
\(\frac{1}{13}+\frac{1}{14}+\frac{1}{15}
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)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{1}{5}+\frac{1}{13}+\frac{1}{14}+\frac{1}{15}+\frac{1}{61}+\frac{1}{62}+\frac{1}{63}
Ta có:
S=1/5+(1/13+1/14+1/15)+(1/61+1/62+1/63)<1/5+1/12.3+1/60.3
=>S<1/5+1/4+1/20=10/20
Hay S<1/2
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Ta có: \(A=\frac{1}{5}+\frac{1}{13}+\frac{1}{14}+\frac{1}{15}+\frac{1}{61}+\frac{1}{62}+\frac{1}{63}\)
\(A=\frac{1}{5}+\left(\frac{1}{13}+\frac{1}{14}+\frac{1}{15}\right)+\left(\frac{1}{62}+\frac{1}{62}+\frac{1}{63}\right)\)
\(A=\frac{1}{5}+\frac{1}{15}.3+\frac{1}{63}.3\)
\(A=\frac{1}{5}+\frac{1}{5}+\frac{1}{21}\)
\(A=\frac{47}{105}\)
Mà: \(\frac{47}{105}< \frac{47}{94}=\frac{1}{2}\)
Nên \(A=\frac{1}{5}+\frac{1}{13}+\frac{1}{14}+\frac{1}{15}+\frac{1}{61}+\frac{1}{62}+\frac{1}{63}< \frac{1}{2}\)
a) Ta có:
S = 1/5 + 1/13 + 1/14 + 1/15 + 1/61 + 1/62 + 1/63
Ta thấy:
1/13 < 1/12 ; 1/14 < 1/12 ; 1/15 < 1/12
=> 1/13 + 1/14 + 1/15 < 1/12 + 1/12 + 1/12 = 1/12 . 3 = 1/4 (1)
1/61 < 1/60 ; 1/62 < 1/60 ; 1/63 < 1/60
=> 1/61 + 1/62 + 1/63 < 1/60 + 1/60 + 1/60 = 1/60. 3 = 1/20 (2)
Từ (1) và (2)
=> 1/13 + 1/14 + 1/15 + 1/61 + 1/62 + 1/63 < 1/4 + 1/20
=>S = 1/5 + 1/13 + 1/14 + 1/15 + 1/61 + 1/62 + 1/63 < 1/4 + 1/20 + 1/5 = 5/20 + 1/20 + 4/20 = 10/20 = 1/2 (ĐPCM)
b) Ta có:
\(P=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{20}}\)
\(2P=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{19}}\)
\(2P-P=1+\frac{1}{2}-\frac{1}{2}+\frac{1}{2^2}-\frac{1}{2^2}+...+\frac{1}{2^{19}}-\frac{1}{2^{19}}-\frac{1}{2^{20}}\)
\(P=1-\frac{1}{2^{20}}< 1\)
=> P < 1