chứng tỏ rằng B= 1/10+1/11+1/12+....+1/99+1/100 B<1 nhanh giúp mk mai thi rồi
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a, Ta có: \(A=\frac{1}{11}+\frac{1}{12}+...+\frac{1}{50}=\left(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{30}\right)+\left(\frac{1}{31}+\frac{1}{32}+...+\frac{1}{60}\right)\)
Nhận xét: \(\frac{1}{11}+\frac{1}{12}+....+\frac{1}{30}>\frac{1}{30}+\frac{1}{30}+...+\frac{1}{30}=\frac{20}{30}=\frac{2}{3}\)
\(\frac{1}{31}+\frac{1}{32}+...+\frac{1}{60}>\frac{1}{60}+\frac{1}{60}+...+\frac{1}{60}=\frac{20}{60}=\frac{1}{3}\)
\(\Rightarrow A>\frac{2}{3}+\frac{1}{3}=1>\frac{1}{2}\)
Vậy A > 1/2
b, Ta có: \(\frac{1}{50}>\frac{1}{100};\frac{1}{51}>\frac{1}{100};........;\frac{1}{99}>\frac{1}{100}\)
\(\Rightarrow B>\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}=\frac{50}{100}=\frac{1}{2}\)
Vậy B > 1/2
c, Ta có: \(C=\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+...+\frac{1}{100}=\frac{1}{10}+\left(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{100}\right)\)
Nhận xét: \(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{100}>\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}=\frac{90}{100}=\frac{9}{10}\)
\(\Rightarrow C>\frac{1}{10}+\frac{9}{10}=\frac{10}{10}=1\)
Vậy C > 1
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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
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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
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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
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Lời giải:
a, Ta có: \(A=\frac{1}{12}+\frac{1}{13}+\frac{1}{14}+\frac{1}{15}+...+\frac{1}{22}>\frac{1}{22}+\frac{1}{22}+\frac{1}{22}+\frac{1}{22}+...+\frac{1}{22}=\frac{1}{22}.11=\frac{11}{22}=\frac{1}{2}\)
Vậy: \(A>\frac{1}{2}\)
b, Ta có: \(B=\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+...+\frac{1}{99}+\frac{1}{100}\)
\(=\left(\frac{1}{10}+\frac{1}{11}+...+\frac{1}{49}+\frac{1}{50}\right)+\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{99}+\frac{1}{100}\right)\)
Mà: \(\left(\frac{1}{10}+\frac{1}{11}+...+\frac{1}{49}+\frac{1}{50}\right)+\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{99}+\frac{1}{100}\right)\text{}\text{}\text{}>\left(\frac{1}{50}+...+\frac{1}{50}+\frac{1}{50}\right)+\left(\frac{1}{100}+...+\frac{1}{100}+\frac{1}{100}\right)\)
=> \(B\text{}\text{}\text{}>\frac{1}{50}.41+\frac{1}{100}.50=\frac{41+25}{50}=\frac{33}{25}>1\)
Vậy: \(B>1\)
c, Ta có: \(C=\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+...+\frac{1}{16}+\frac{1}{17}< \frac{1}{5}+\frac{1}{6}+\left(\frac{1}{7}+...+\frac{1}{7}+\frac{1}{7}\right)=\frac{11}{30}+11.\frac{1}{7}=\frac{407}{210}< \frac{420}{210}=2\)
Vậy: \(C< 2\)
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