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ta có: \(A=\left(\frac{1}{11}+...+\frac{1}{20}\right)+\left(\frac{1}{21}+...+\frac{1}{30}\right)+\left(\frac{1}{31}+...+\frac{1}{60}\right)+...+\frac{1}{70}\)
mà \(\frac{1}{11}+...+\frac{1}{20}>\frac{1}{20}+...+\frac{1}{20}=\frac{10}{20}=\frac{1}{2}\)
\(\frac{1}{21}+...+\frac{1}{30}>\frac{1}{30}+...+\frac{1}{30}=\frac{10}{30}=\frac{1}{3}\)
\(\frac{1}{31}+...+\frac{1}{60}>\frac{1}{60}+...+\frac{1}{60}=\frac{30}{60}=\frac{1}{2}\)
\(\Rightarrow A>\frac{1}{2}+\frac{1}{3}+\frac{1}{2}+\frac{1}{61}+...+\frac{1}{70}>\frac{1}{2}+\frac{1}{3}+\frac{1}{2}=\frac{4}{3}\)
\(\Rightarrow A>\frac{4}{3}\left(1\right)\)
ta có: \(A=\left(\frac{1}{11}+...+\frac{1}{20}\right)+\left(\frac{1}{21}+...+\frac{1}{30}\right)+\left(\frac{1}{31}+...+\frac{1}{40}\right)+\left(\frac{1}{41}+...+\frac{1}{50}\right)\)
\(+\left(\frac{1}{51}+...+\frac{1}{60}\right)+\left(\frac{1}{61}+...+\frac{1}{70}\right)\)
mà \(\frac{1}{11}+...+\frac{1}{20}< \frac{1}{11}+...+\frac{1}{11}=\frac{10}{11}< \frac{10}{10}=1\)
\(\frac{1}{21}+...+\frac{1}{30}< \frac{1}{21}+...+\frac{1}{21}=\frac{10}{21}< \frac{10}{20}=\frac{1}{2}\)
\(\frac{1}{31}+...+\frac{1}{40}< \frac{1}{31}+...+\frac{1}{31}=\frac{10}{31}< \frac{10}{30}=\frac{1}{3}\)
\(\frac{1}{41}+...+\frac{1}{50}< \frac{1}{41}+...+\frac{1}{41}=\frac{10}{41}< \frac{10}{40}=\frac{1}{4}\)
\(\frac{1}{51}+...+\frac{1}{60}< \frac{1}{51}+...+\frac{1}{51}=\frac{10}{51}< \frac{10}{50}=\frac{1}{5}\)
\(\frac{1}{61}+...+\frac{1}{70}< \frac{1}{61}+...+\frac{1}{61}=\frac{10}{61}< \frac{10}{60}=\frac{1}{6}\)
\(\Rightarrow A< 1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}=1+\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{6}\right)+\left(\frac{1}{4}+\frac{1}{5}\right)\)
\(=1+1+\frac{9}{20}< 1+1+\frac{10}{20}=\frac{5}{2}=2,5\)
\(\Rightarrow A< 2,5\left(2\right)\)
từ (1); (2) \(\Rightarrow\frac{4}{3}< A< 2,5\left(đpcm\right)\)
CHÚC BN HỌC TỐT!
\(A=\left(\frac{1}{11}+....+\frac{1}{20}\right)+\left(\frac{1}{21}+....+\frac{1}{30}\right)+\left(\frac{1}{31}+....+\frac{1}{40}\right)+\left(\frac{1}{41}+....+\frac{1}{50}\right)\)
\(+\left(\frac{1}{51}+....+\frac{1}{60}\right)+\left(\frac{1}{61}+....+\frac{1}{70}\right)\)
\(A
\(A=\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+...+\frac{1}{70}\)
\(\Rightarrow A>\frac{1}{70}+\frac{1}{70}+\frac{1}{70}+...+\frac{1}{70}\)(60 số hạng)
\(\Rightarrow A>\frac{60}{70}>\frac{60}{80}=\frac{3}{4}\)
Vậy \(A>\frac{3}{4}\left(đpcm\right)\)
\(A=\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+...+\frac{1}{70}\)
\(\Rightarrow A>\frac{1}{70}+...+\frac{1}{70}\)(60 số hạng)
\(\Rightarrow A>\frac{60}{70}>\frac{60}{60}=\frac{3}{4}\)
ta có số hạng là 60 số hạng
nếu có 5 nhóm thì mỗi nhóm có 12 số hạng
=(1/11+1/12+.....+1/21+1/22)+(1/23+1/24+...+1/33+1/34)+(1/35+1/36+...+1/45+1/46)+ (1/47+1/48+....+1/56+1/57)+(1/58+1/59+1/69+1/70)
xét nhóm 1 ta có
1/11=1/11
1/11>1/12
1/11>1/13
................
1/11>1/22
xét nhóm 2 ta có
1/23=1/23
1/23>1/24
1/23>1/25
................
1/23>1/34
Xét nhóm 3 ta có
1/35=1/35
1/35>1/36
................
1/35>1/46
Xét nhóm 4 ta có
1/47=1/47
1/47>1/48
.................
1/47>1/57
Xét nhóm 5 ta có
1/58=1/58
1/58>1/59
................
1/58>1/70
Vây ta có A<1/11.12+1/23.12+1/35.12+1/47.12+1/58.12
Ta có 1/11.12+1/23.12+1/35.12+1/47.12+1/58.12<5/2
Dựa vào tính chất bắc cầu thì A<5/2
Vẫn chia 5 nhóm ta có
nhóm 1
1/11>1/22
1/12>1/22
................
1/22=1/22
Xét nhóm 2 ta có
1/23>1/34
1/24>1/34
................
1/34=1/34
Xét nhóm 3 ta có
1/35>1/46
1/34>1/46
................
1/46=1/46
Xét nhóm 4 ta có
1/47>1/57
1/48>1/57
................
1/57=1/57
Xét nhóm 5 ta có
1/58>1/70
1/59>1/70
...............
1/70=1/70
Vậy ta có A>1/22.12+1/34.12+1/46.12+1/57.12+1/70.12
mà 1/22.12+1/34.12+1/46.12+1/57.12+1/70.12>4/3
Vậy A>4/3
Vậy 4/3<A<5/2
\(A=\frac{1}{11}+\frac{1}{12}+...+\frac{1}{70}.\)
\(\Rightarrow A=\left(\frac{1}{11}+\frac{1}{12}+..+\frac{1}{20}\right)+\left(\frac{1}{21}+...+\frac{1}{30}\right)+\left(\frac{1}{31}+..+\frac{1}{60}\right)+..+\frac{1}{70}\)
Ta có :
\(\frac{1}{10}+...+\frac{1}{10}=1>\frac{1}{11}+\frac{1}{12}+..+\frac{1}{20}>\frac{1}{20}+..+\frac{1}{20}=\frac{10}{20}=\frac{1}{2}\)
\(\frac{1}{20}+..+\frac{1}{20}=\frac{10}{20}=\frac{1}{2}>\frac{1}{21}+\frac{1}{22}+...+\frac{1}{30}>\frac{1}{30}+\frac{1}{30}+...+\frac{1}{30}=\frac{10}{30}=\frac{1}{3}\)
\(\frac{1}{30}+..+\frac{1}{30}=\frac{30}{30}=1>\frac{1}{31}+\frac{1}{32}+..+\frac{1}{60}>\frac{1}{60}+\frac{1}{60}+..+\frac{1}{60}=\frac{30}{60}=\frac{1}{2}\)
\(\Rightarrow1+1+\frac{1}{2}=\frac{5}{2}>A=\left(\frac{1}{11}+..+\frac{1}{20}\right)+\left(\frac{1}{21}+...+\frac{1}{30}\right)+\left(\frac{1}{31}+..+\frac{1}{60}\right)+..+\frac{1}{70}>\)
\(\frac{1}{2}+\frac{1}{2}+\frac{1}{2}=\frac{4}{3}\)
\(\Rightarrow\frac{5}{2}>A>\frac{4}{3}\)