Giải:
Đặt \(A=\dfrac{1}{31}+\dfrac{1}{32}+\dfrac{1}{33}+...+\dfrac{1}{59}+\dfrac{1}{60}\)
Ta có:
\(A=\dfrac{1}{31}+\dfrac{1}{32}+\dfrac{1}{33}+...+\dfrac{1}{59}+\dfrac{1}{60}\)
\(\Rightarrow A=\left(\dfrac{1}{31}+\dfrac{1}{32}+...+\dfrac{1}{40}\right)+\left(\dfrac{1}{41}+\dfrac{1}{42}+...+\dfrac{1}{50}\right)+\left(\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{60}\right)\)
Nhận xét:
\(\dfrac{1}{31}+\dfrac{1}{32}+...+\dfrac{1}{40}< \dfrac{1}{30}+\dfrac{1}{30}+...+\dfrac{1}{30}=\dfrac{1}{3}\)
\(\dfrac{1}{41}+\dfrac{1}{42}+...+\dfrac{1}{50}< \dfrac{1}{40}+\dfrac{1}{40}+...+\dfrac{1}{40}=\dfrac{1}{4}\)
\(\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{60}< \dfrac{1}{50}+\dfrac{1}{50}+...+\dfrac{1}{50}=\dfrac{1}{5}\)
\(\Rightarrow A< \dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}=\dfrac{47}{60}< \dfrac{48}{60}=\dfrac{4}{5}\)
\(\Rightarrow A< \dfrac{4}{5}\left(1\right)\)
Lại có:
\(\dfrac{1}{31}+\dfrac{1}{32}+...+\dfrac{1}{40}>\dfrac{1}{40}+\dfrac{1}{40}+...+\dfrac{1}{40}=\dfrac{1}{4}\)
\(\dfrac{1}{41}+\dfrac{1}{42}+...+\dfrac{1}{50}>\dfrac{1}{50}+\dfrac{1}{50}+...+\dfrac{1}{50}=\dfrac{1}{5}\)
\(\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{60}>\dfrac{1}{60}+\dfrac{1}{60}+...+\dfrac{1}{60}=\dfrac{1}{6}\)
\(\Rightarrow A>\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}=\dfrac{37}{60}>\dfrac{36}{60}=\dfrac{3}{5}\)
\(\Rightarrow A>\dfrac{3}{5}\left(2\right)\)
Từ \(\left(1\right)\) và \(\left(2\right)\)
\(\Rightarrow\dfrac{3}{5}< A< \dfrac{4}{5}\)
Vậy \(\dfrac{3}{5}< \dfrac{1}{31}+\dfrac{1}{32}+\dfrac{1}{33}+...+\dfrac{1}{59}+\dfrac{1}{60}< \dfrac{4}{5}\) (Đpcm)
Đặt A=131+132+133+...+159+160A=131+132+133+...+159+160
Ta có:
A=131+132+133+...+159+160A=131+132+133+...+159+160
⇒A=(131+132+...+140)+(141+142+...+150)+(151+152+...+160)⇒A=(131+132+...+140)+(141+142+...+150)+(151+152+...+160)
Nhận xét:
131+132+...+140<130+130+...+130=13131+132+...+140<130+130+...+130=13
141+142+...+150<140+140+...+140=14141+142+...+150<140+140+...+140=14
151+152+...+160<150+150+...+150=15151+152+...+160<150+150+...+150=15
⇒A<13+14+15=4760<4860=45⇒A<13+14+15=4760<4860=45
⇒A<45(1)⇒A<45(1)
Lại có:
131+132+...+140>140+140+...+140=14131+132+...+140>140+140+...+140=14
141+142+...+150>150+150+...+150=15141+142+...+150>150+150+...+150=15
151+152+...+160>160+160+...+160=16151+152+...+160>160+160+...+160=16
⇒A>14+15+16=3760>3660=35⇒A>14+15+16=3760>3660=35
⇒A>35(2)⇒A>35(2)
Từ (1)(1) và (2)(2)
⇒35<A<45⇒35<A<45
Vậy 35<131+132+133+...+159+160<4535<131+132+133+...+159+160<45
