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đề sai hả bạn số hạng cuối có phải là \(\frac{1}{100}\)đúng không
Ta có :
A = \(\dfrac{1}{10}\) + \(\dfrac{1}{11}\) + \(\dfrac{1}{12}\) +.................+ \(\dfrac{1}{99}\) + \(\dfrac{1}{100}\) ( 91 số hạng)
A = \(\dfrac{1}{10}\) + \(\left(\dfrac{1}{11}+\dfrac{1}{12}+...........+\dfrac{1}{99}+\dfrac{1}{100}\right)\)
Vì \(\dfrac{1}{11}>\dfrac{1}{100}\)
\(\dfrac{1}{12}>\dfrac{1}{100}\)
.................................
\(\dfrac{1}{99}< \dfrac{1}{100}\)
\(=>\) \(A\) > \(\dfrac{1}{10}+\left(\dfrac{1}{100}+\dfrac{1}{100}+........+\dfrac{1}{100}\right)\) (90 số hạng \(\dfrac{1}{100}\) )
A > \(\dfrac{1}{10}+\dfrac{90}{100}\)
\(A\) > \(\dfrac{1}{10}+\dfrac{9}{10}\)
=> A > 1
=> đpcm
\(\frac{1}{10}+\frac{1}{11}+...+\frac{1}{19}>\frac{1}{20}+\frac{1}{20}+...+\frac{1}{20}=\frac{10}{20}=\frac{1}{2}\)
\(\frac{1}{20}+\frac{1}{21}+...+\frac{1}{29}>\frac{1}{30}+\frac{1}{30}+...+\frac{1}{30}=\frac{10}{30}=\frac{1}{3}\)
\(\frac{1}{30}+\frac{1}{31}+...+\frac{1}{39}>\frac{1}{40}+\frac{1}{40}+...+\frac{1}{40}=\frac{10}{40}=\frac{1}{4}\)
\(\Rightarrow\frac{1}{10}+\frac{1}{11}+...+\frac{1}{39}>\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\)
\(\frac{13}{12}\) \(>\) \(1\)
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
C = 1 / 10 + ( 1 / 11 + 1 / 12 + ... + 1 / 99 + 1 / 100 )
<=> 1 / 10 + ( 1 / 11 + 1 / 12 + ... + 1 / 99 + 1 / 100 ) > 1 / 10 + ( 1 / 100 + 1 / 100 + ... + 1 / 100 )
<=> 1/ 10 + 90 / 100 = 1
Vậy C > 1 (đpcm)
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
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
C = 1 10 + 1 11 + 1 12 + ... + 1 99 + 1 100 > 1 ⇔ C = 1 10 + 1 11 + 1 12 + ... + 1 20 + 1 21 + 1 22 + ... + 1 30 + ... + 1 91 + 1 92 + ... + 1 100 C > 1 10 + 10 20 + 10 30 + ... + 10 100 > 10 20 + 10 30 + 10 60 = 1 2 + 1 3 + 1 6 = 1