Cho A= 1/31+1/32+1/33+.....+1/60
Chứng minh 3/5<A<4/5
GIẢI
Ta có :
\(A=\frac{1}{31}+\frac{1}{32}+\frac{1}{33}+.........+\frac{1}{60}\)
\(\Leftrightarrow A=\left(\frac{1}{31}+\frac{1}{32}+....+\frac{1}{40}\right)+\left(\frac{1}{41}+\frac{1}{42}+....+\frac{1}{50}\right)+\left(\frac{1}{51}+\frac{1}{52}+.....+\frac{1}{60}\right)\left(1\right)\)
Mà:
\(\frac{1}{31}>\frac{1}{32}>\frac{1}{33}>\frac{1}{34}>\frac{1}{35}>\frac{1}{36}>\frac{1}{37}>\frac{1}{38}>\frac{1}{39}>\frac{1}{40}\)
\(\Rightarrow\frac{1}{31}+\frac{1}{32}+........+\frac{1}{40}>\frac{1}{40}+.......+\frac{1}{40}\)
\(\Leftrightarrow\frac{1}{31}+\frac{1}{32}+......+\frac{1}{40}>10\times\frac{1}{40}\)
\(\Leftrightarrow\frac{1}{31}+\frac{1}{32}+..........+\frac{1}{40}>\frac{1}{4}\)
Tương tự:
\(\frac{1}{41}+\frac{1}{42}+.........+\frac{1}{50}>\frac{1}{5}\)
\(\frac{1}{51}+\frac{1}{52}+.....+\frac{1}{60}>\frac{1}{6}\)
\(\Rightarrow A>\frac{1}{4}+\frac{1}{5}+\frac{1}{6}=\frac{37}{60}>\frac{36}{60}=\frac{3}{5}\)
Vậy \(\frac{3}{5}< A\left(2\right)\)
Từ (1), ta lại có:
\(\frac{1}{31}+\frac{1}{32}+.......+\frac{1}{40}< 10\times\frac{1}{30}=\frac{1}{3}\)
\(\frac{1}{41}+\frac{1}{42}+..........+\frac{1}{50}< 10\times\frac{1}{40}=\frac{1}{4}\)
\(\frac{1}{51}+\frac{1}{52}+.........+\frac{1}{60}< 10\times\frac{1}{50}=\frac{1}{5}\)
\(\Rightarrow A< \frac{1}{3}+\frac{1}{4}+\frac{1}{5}=\frac{47}{60}< \frac{48}{60}=\frac{4}{5}\)
Vậy \(A< \frac{4}{5}\left(3\right)\)
Từ (2) và (3) , suy ra:
\(\frac{3}{5}< A< \frac{4}{5}\)
ad ơi cho em hỏi là tại sao lại phải nhóm 10 phân số 1 nhóm vậy ạk
