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1/10+1/11+…+1/19 > 1/20+1/20+…+1/20 = 10/20 = 1/2
1/20+1/21+…+1/29 > 1/30+1/30+…+1/30 = 10/30 = 1/3
1/30+1/31+…+1/39 > 1/40+1/40+…+1/40 = 10/40 = 1/4
=> A>1
Chỉ cần 30 số hạng đầu đã lớn hơn 1.
1/10+1/11+…+1/19 > 1/20+1/20+…+1/20 = 10/20 = 1/2
1/20+1/21+…+1/29 > 1/30+1/30+…+1/30 = 10/30 = 1/3
1/30+1/31+…+1/39 > 1/40+1/40+…+1/40 = 10/40 = 1/4
=>
1/10+1/11+…+1/39 > 1/2+1/3+1/4 = 13/12 > 1
A = 1 / 10 + ( 1 / 11 + 1 / 12 + ... + 1 / 99 + 1 / 100 )
A = 1 / 10 + ( 1 / 11 + 1 / 12 + ... + 1 / 99 + 1 / 100 ) > 1 / 10 + ( 1 / 100 + 1 / 100 + ... + 1 / 100 )
= 1 / 10 + 90 / 100 = 1
Vậy A > 1
1/10+1/11+…+1/19 > 1/20+1/20+…+1/20 = 10/20 = 1/2
1/20+1/21+…+1/29 > 1/30+1/30+…+1/30 = 10/30 = 1/3
1/30+1/31+…+1/39 > 1/40+1/40+…+1/40 = 10/40 = 1/4
=>
1/10+1/11+…+1/39 > 1/2+1/3+1/4 = 13/12 > 1
đúng nhé
Ta có :
Cần 30 số hạng đầu đã lớn hơn 1.
1/10+1/11+…+1/19 > 1/20+1/20+…+1/20 = 10/20 = 1/2
1/20+1/21+…+1/29 > 1/30+1/30+…+1/30 = 10/30 = 1/3
1/30+1/31+…+1/39 > 1/40+1/40+…+1/40 = 10/40 = 1/4
=> 1/10+1/11+…+1/39 > 1/2+1/3+1/4 = 13/12 > 1
Vậy :C>1
a: Ta có
A = \(\dfrac{1}{10}\) + \((\dfrac{1}{11}\) + \(\dfrac{1}{12}\) + ...+ \(\dfrac{1}{100}\)\()\)
⇒ A > \(\dfrac{1}{10}\) + \((\dfrac{1}{100}\) + \(\dfrac{1}{100}\) + ...+ \(\dfrac{1}{100}\)\()\)90 số hạng
⇒ A > \(\dfrac{1}{10}\) + \(\dfrac{90}{100}\)
⇒ A > 1
vậy A > 1
b: ta có
S = (\(\dfrac{1}{21}\) + \(\dfrac{1}{22}\)+ \(\dfrac{1}{23}\) + \(\dfrac{1}{24}\) + \(\dfrac{1}{25}\))+(\(\dfrac{1}{26}\) + \(\dfrac{1}{27}\)+ \(\dfrac{1}{28}\) + \(\dfrac{1}{29}\) + \(\dfrac{1}{30}\))+(\(\dfrac{1}{31}\) + \(\dfrac{1}{32}\)+ \(\dfrac{1}{33}\) + \(\dfrac{1}{34}\) + \(\dfrac{1}{35}\))
⇒ S > (\(\dfrac{1}{25}\) + \(\dfrac{1}{25}\)+ \(\dfrac{1}{25}\) + \(\dfrac{1}{25}\) + \(\dfrac{1}{25}\))+(\(\dfrac{1}{30}\) + \(\dfrac{1}{30}\)+ \(\dfrac{1}{30}\) + \(\dfrac{1}{30}\) + \(\dfrac{1}{30}\))+(\(\dfrac{1}{35}\) + \(\dfrac{1}{35}\)+ \(\dfrac{1}{35}\) + \(\dfrac{1}{35}\) + \(\dfrac{1}{35}\))
⇔ S > \(\dfrac{5}{25}\)+\(\dfrac{5}{30}\)+\(\dfrac{5}{35}\)
⇔ S > \(\dfrac{1}{5}\)+\(\dfrac{1}{6}\)+\(\dfrac{1}{7}\)
⇔ S > \(\dfrac{107}{210}\)> \(\dfrac{105}{210}\)=\(\dfrac{1}{2}\)
vậy S > \(\dfrac{1}{2}\)
1/10>1/40;1/11>1/40;...;1/40=1/40
=>1/10+1/11+...+1/40>1/40+1/40+...+1/=1/40 .31=31/40>1/2 (1)
1/41>1/100;1/42>1/100;....1/100=1/100
=>1/41+1/42+...+1/100>1/100+1/100+...+1/100=1/100 .60=3/5>1/2 (2)
Từ (1) và (2) =>A>1/2+1/2=1
Vậy A>1(đpcm)
A = 1/10+ (1/11 + 1/12+ ...+ 1/99 + 1/100)
A = 1/10 + (1/11 + 1/12 + ..+ 1/99 + 1/100) > 1/10 + (1/100 + 1/100 +... + 1/100)
A = 1/10 + 90/100 = 1
vậy A > 1
30 số hạng đầu lớn hơn 1
\(\frac{1}{10}+\frac{1}{11}+..+\frac{1}{19}>\frac{1}{20}+\frac{1}{20}+..+\frac{1}{20}=\frac{1}{2}\)\(\frac{1}{2}\)
\(\frac{1}{20}+\frac{1}{21}+..+\frac{1}{29}>\frac{1}{30}+\frac{1}{30}+..+\frac{1}{30}=\frac{1}{3}\)
\(\frac{1}{30}+\frac{1}{31}+...+\frac{1}{39}>\frac{1}{40}+\frac{1}{40}+...+\frac{1}{40}=\frac{1}{4}\)
=> \(\frac{1}{10}+\frac{1}{11}+...+\frac{1}{39}>\frac{1}{2}+\frac{1}{3}+\frac{1}{4}=\frac{13}{12}>1\)
\(A=\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+...+\frac{1}{99}+\frac{1}{100}\)
\(A=\frac{1}{10}+\left(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{99}+\frac{1}{100}\right)>\frac{1}{10}+\left(\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}\right)\)
\(A=\frac{1}{10}+\frac{99}{100}=1\)
=> A > 1
\(A=\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+...+\frac{1}{99}+\frac{1}{100}\)
\(A=\frac{1}{10}+\frac{1}{11}+...+\frac{1}{19}>\frac{1}{20}+\frac{1}{20}+...+\frac{1}{20}=\frac{10}{20}=\frac{1}{2}\)
\(A=\frac{1}{20}+\frac{1}{21}+...+\frac{1}{29}>\frac{1}{30}+\frac{1}{30}+...+\frac{1}{30}=\frac{10}{30}=\frac{1}{3}\)
\(A=\frac{1}{30}+\frac{1}{31}+...+\frac{1}{39}>\frac{1}{40}+\frac{1}{40}+... +\frac{1}{40}=\frac{10}{40}=\frac{1}{4}\)
\(\Rightarrow A>1\)
A = 1 10 + 1 11 + 1 12 + ... + 1 99 + 1 100 = 1 10 + 1 11 + 1 12 + ... + 1 90 + 1 91 + 1 92 + ... 1 99 + 1 100
Đặt A 1 = 1 10 + 1 11 + 1 12 + ... + 1 90 ; A 2 = 1 91 + 1 92 + ... 1 99 + 1 100
Ta có 1 10 > 1 90 ; 1 11 > 1 90 ; 1 12 > 1 90 ; ... ; 1 89 > 1 90
A 1 = 1 10 + 1 11 + 1 12 + ... + 1 90 > 81 ⋅ 1 90
A 1 = 1 10 + 1 11 + 1 12 + ... + 1 90 > 81 ⋅ 1 90
Tương tự 1 91 > 1 100 ; 1 92 > 1 100 ; ... ; 1 99 > 1 100
A 2 = 1 91 + 1 92 + ... 1 99 + 1 100 > 10 ⋅ 1 100
A 2 = 1 91 + 1 92 + ... 1 99 + 1 100 > 10 100 = 1 10
A = 1 10 + 1 11 + 1 12 + ... + 1 99 + 1 100 = A 1 + A 2 > 9 10 + 1 10
A = 1 10 + 1 11 + 1 12 + ... + 1 99 + 1 100 > 1