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vi 1/62>1/80 ;1/62>1/80:...:1/80=0/80
suy ra 1/61+1/62+1/63+...+1/80>1/80+1/80+1/80+...+1/80
moi ve co 20 so hang
Vì \(\frac{1}{13}+\frac{1}{14}+\frac{1}{15}<\frac{1}{12}+\frac{1}{12}+\frac{1}{12}=\frac{3}{12}=\frac{1}{4}\)
\(\frac{1}{61}+\frac{1}{62}+...+\frac{1}{66}<\frac{1}{60}+\frac{1}{60}+...+\frac{1}{60}=\frac{6}{60}=\frac{1}{10}\)
=> A < \(\frac{1}{3}+\frac{1}{4}+\frac{1}{10}=\frac{41}{60}<\frac{45}{60}=\frac{3}{4}\)điều phải c/m
Ta có :
\(S=\frac{1}{17}+\frac{1}{18}+\frac{1}{19}+.......+\frac{1}{62}+\frac{1}{63}+\frac{1}{64}\)
\(\Rightarrow S< \frac{1}{17}+\frac{1}{17}+......+\frac{1}{17}+\frac{1}{17}+\frac{1}{17}\)
\(\Rightarrow S< \frac{1}{17}.48\)
\(\Rightarrow S< \frac{48}{17}\)
\(\Rightarrow S< 2\)( 1 )
Lại có :
\(S>\frac{1}{64}+\frac{1}{64}+.........+\frac{1}{64}+\frac{1}{64}+\frac{1}{64}\)
\(\Rightarrow S>\frac{1}{64}.48\)
\(\Rightarrow S>\frac{3}{4}\)( 2 )
Từ ( 1 ) và ( 2 ) suy ra : \(\frac{3}{4}< S< 2\)
Vậy \(1< S< 2\left(ĐPCM\right)\)
\(\frac{1}{5}+\frac{1}{13}+\frac{1}{14}+\frac{1}{15}+\frac{1}{61}+\frac{1}{62}+\frac{1}{63}<\frac{1}{5}+\frac{1}{13}.3+\frac{1}{61}.3\)
\(=\frac{1}{5}+\frac{3}{13}+\frac{3}{61}<\frac{1}{5}+\frac{3}{12}+\frac{3}{60}=\frac{1}{5}+\frac{1}{4}+\frac{1}{20}=\frac{1}{2}\)
\(\Rightarrowđpcm\)
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
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
TA có:
1/12>1/13
1/12>1/14
1/12>1/15
=>1/12.3=1/4>1/13+1/14+1/15
1/60>1/61
1/60>1/62
1/60>1/63
=>1/60.3=1/20>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
=>1/2> 1/5+1/13+1/14+1/15+1/61+1/62+1/63