Cho b=1/61+1/62+1/63+1/64+...+1/119+1/120. Chứng minh rằng 7/12<b<1
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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
Đặt :
\(A=\)\(\dfrac{1}{5}+\dfrac{1}{13}+\dfrac{1}{14}+\dfrac{1}{15}+\dfrac{1}{61}+\dfrac{1}{62}+\dfrac{1}{63}\)
\(A=\dfrac{1}{5}+\left(\dfrac{1}{13}+\dfrac{1}{14}+\dfrac{1}{15}\right)+\left(\dfrac{1}{61}+\dfrac{1}{62}+\dfrac{1}{63}\right)\)
Ta thấy :
\(\dfrac{1}{13}+\dfrac{1}{14}+\dfrac{1}{15}< \dfrac{1}{12}+\dfrac{1}{12}+\dfrac{1}{12}\)
\(\dfrac{1}{61}+\dfrac{1}{62}+\dfrac{1}{63}< \dfrac{1}{60}+\dfrac{1}{61}+\dfrac{1}{62}\)
\(\Rightarrow A< \dfrac{1}{5}+\left(\dfrac{1}{12}+\dfrac{1}{12}+\dfrac{1}{12}\right)+\left(\dfrac{1}{60}+\dfrac{1}{60}+\dfrac{1}{60}\right)\)
\(\Rightarrow A< \dfrac{1}{5}+\dfrac{1}{12}.3+\dfrac{1}{60}.3\)
\(\Rightarrow A< \dfrac{1}{5}+\dfrac{1}{4}+\dfrac{1}{20}\)
\(\Rightarrow A< \dfrac{10}{20}=\dfrac{1}{2}\)
\(\Rightarrow A< \dfrac{1}{2}\rightarrowđpcm\)
\(\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
1 - 2 - 3 + 4 + 5 - 6 - 7 + 8 + 9 - 10 - 11 + 12 + ... + 61 - 62 - 63 + 64 ( 64 số )
= ( 1 - 2 - 3 + 4 ) + ( 5 - 6 - 7 + 8 ) + ( 9 - 10 - 11 + 12 ) + ... + ( 61 - 62 - 63 + 64 ) ( 16 nhóm )
= 0 + 0 + 0 + ... + 0 ( 16 số 0 )
= 0 . 16
= 0