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Sửa đề: \(S=\dfrac{1}{20}+\dfrac{1}{21}+\dfrac{1}{22}+...+\dfrac{1}{50}\)
Ta có: \(S=\dfrac{1}{20}+\dfrac{1}{21}+\dfrac{1}{22}+...+\dfrac{1}{50}\)
\(=\dfrac{1}{20}+\left(\dfrac{1}{21}+\dfrac{1}{22}+...+\dfrac{1}{30}\right)+\left(\dfrac{1}{31}+\dfrac{1}{32}+...+\dfrac{1}{40}\right)+\left(\dfrac{1}{41}+\dfrac{1}{42}+...+\dfrac{1}{50}\right)\)
\(\Leftrightarrow S>\dfrac{1}{20}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}=\dfrac{1}{4}+\dfrac{1}{3}+\dfrac{1}{4}\)
\(\Leftrightarrow S>\dfrac{1}{4}+\dfrac{1}{4}+\dfrac{1}{4}=\dfrac{3}{4}\)(đpcm)
Có : \(S=1+2+2^2+2^3+....+2^{99}\)
\(\Rightarrow2S=2+2^2+2^3+....+2^{100}\)
\(\Rightarrow2S-S=\left(2+2^2+2^3+...+2^{100}\right)-\left(1+2+2^2+....+2^{99}\right)\)
\(\Rightarrow S=2^{100}-1< 2^{100}\)
Vậy \(S< 2^{100}\)
S=1+2+22+23+....+299
⇒2S=2+22+23+....+2100
⇒2S−S=2100-1
S=2100-1
vì 2100 -1<2100
⇒S<2100
ta có 1/3=10/30
1/21+1/22+...+1/30 có 10 p/số
mà 1/21>1/30
1/22>1/30
....
1/29>1/30
1/30=1/30
=>1/21+..1/30>1/30+....1/30 có 10 phân số
=>1/21+...1/30>1/3
do \(\frac{5}{20}< 1;\frac{5}{21}< 1;\frac{5}{22}< 1;\frac{5}{23}< 1;\frac{5}{24}< 1\)
\(\Rightarrow\frac{5}{20}+\frac{5}{21}+\frac{5}{22}+\frac{5}{23}+\frac{5}{24}< 1\)
Vậy S < 1
Mk nghĩ thế bn ạ
Ai thấy tớ đúng ủng hộ nha
Ta có :
\(\frac{5}{20}>\frac{5}{25}\)
\(\frac{5}{21}>\frac{5}{25}\)
\(\frac{5}{22}>\frac{5}{25}\)
\(\frac{5}{23}>\frac{5}{25}\)
\(\frac{5}{24}>\frac{5}{25}\)
\(\Rightarrow\frac{5}{20}+\frac{5}{21}+\frac{5}{22}+\frac{5}{23}+\frac{5}{24}>5.\frac{5}{25}=1\)
\(\Rightarrow\frac{5}{20}+\frac{5}{21}+\frac{5}{22}+\frac{5}{23}+\frac{5}{24}>1\)
ta có S=5/20+5/21+5/22+5/23+5/24>5/25+5/25+5/25+5/25+5/25=5/25*5=1
=>đpcm
\(S=\frac{1}{21}+\frac{1}{22}+...+\frac{1}{150}\)
\(=\left(\frac{1}{21}+...+\frac{1}{40}\right)+\left(\frac{1}{41}+...+\frac{1}{80}\right)+\left(\frac{1}{81}+...+\frac{1}{150}\right)\)
\(>\left(\frac{1}{40}+...+\frac{1}{40}\right)+\left(\frac{1}{80}+...+\frac{1}{80}\right)+\left(\frac{1}{150}+...+\frac{1}{150}\right)\)
\(=\frac{20}{40}+\frac{40}{80}+\frac{70}{150}\)
\(=\frac{1}{2}+\frac{1}{2}+\frac{7}{15}>\frac{5}{4}\)