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NV
23 tháng 2 2021

\(\dfrac{1}{1+2+3+...+n}=\dfrac{1}{\dfrac{n\left(n+1\right)}{2}}=\dfrac{2}{n\left(n+1\right)}=\dfrac{2}{n}-\dfrac{2}{n+1}\)

Do đó:

\(\dfrac{1}{1+2+3}+\dfrac{1}{1+2+3+4}+...+\dfrac{1}{1+2+...+59}=\dfrac{2}{3}-\dfrac{2}{4}+\dfrac{2}{4}-\dfrac{2}{5}+...+\dfrac{2}{59}-\dfrac{2}{60}\)

\(=\dfrac{2}{3}-\dfrac{2}{60}< \dfrac{2}{3}\) (đpcm)

9 tháng 2 2018

\(M=\dfrac{1}{1+2+3}+\dfrac{1}{1+2+3+4}+...+\dfrac{1}{1+2+3+...+59}\\ =\dfrac{1}{\dfrac{3\cdot4}{2}}+\dfrac{1}{\dfrac{4\cdot5}{2}}+...+\dfrac{1}{\dfrac{59\cdot60}{2}}\\ =\dfrac{2}{3\cdot4}+\dfrac{2}{4\cdot5}+...+\dfrac{2}{59\cdot60}\\ =2\left(\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+...+\dfrac{1}{59\cdot60}\right)\\ =2\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{59}-\dfrac{1}{60}\right)\\ =2\cdot\dfrac{19}{60}\\ =\dfrac{38}{60}< \dfrac{40}{60}=\dfrac{2}{3}\)

25 tháng 3 2017

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 \(\left(1\right)\)\(\left(2\right)\)

\(\Rightarrow\dfrac{3}{5}< A< \dfrac{4}{5}\)

Vậy \(\dfrac{3}{5}< \dfrac{1}{31}+\dfrac{1}{32}+\dfrac{1}{33}+...+\dfrac{1}{59}+\dfrac{1}{60}< \dfrac{4}{5}\) (Đpcm)

22 tháng 4 2018

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)(2)(2)

⇒35<A<45⇒35<A<45

Vậy 35<131+132+133+...+159+160<4535<131+132+133+...+159+160<45

14 tháng 11 2023

2:

\(B=\left(\dfrac{1}{2^2}-1\right)\left(\dfrac{1}{3^2}-1\right)\cdot...\cdot\left(\dfrac{1}{100^2}-1\right)\)

\(=\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{2}+1\right)\left(\dfrac{1}{3}-1\right)\left(\dfrac{1}{3}+1\right)\cdot...\cdot\left(\dfrac{1}{100}-1\right)\left(\dfrac{1}{100}+1\right)\)

\(=\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right)\cdot...\cdot\left(\dfrac{1}{100}-1\right)\left(\dfrac{1}{2}+1\right)\left(\dfrac{1}{3}+1\right)\cdot...\cdot\left(\dfrac{1}{100}+1\right)\)

\(=\dfrac{-1}{2}\cdot\dfrac{-2}{3}\cdot...\cdot\dfrac{-99}{100}\cdot\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{101}{100}\)

\(=-\dfrac{1}{100}\cdot\dfrac{101}{2}=\dfrac{-101}{200}< -\dfrac{100}{200}=-\dfrac{1}{2}\)

 

18 tháng 4 2023

Ta có \(M=\dfrac{1}{1+2+3}+\dfrac{1}{1+2+3+4}+...+\dfrac{1}{1+2+3+...+59}\)

              = \(\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+...+\dfrac{1}{59\cdot60}\)

              = \(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{59}-\dfrac{1}{60}\)

              = \(\dfrac{1}{3}-\dfrac{1}{60}=\dfrac{19}{60}< \dfrac{40}{60}=\dfrac{2}{3}\)

Vậy M < \(\dfrac{2}{3}\)

18 tháng 4 2023

Ta có: 

loading...

AH
Akai Haruma
Giáo viên
30 tháng 4 2023

Lời giải:

$A< \frac{1}{2^2}+\frac{1}{2.3}+\frac{1}{3.4}+..+\frac{1}{99.100}$

$A< \frac{1}{4}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{100-99}{99.100}$

$A< \frac{1}{4}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}$

$A< \frac{1}{4}+\frac{1}{2}-\frac{1}{100}< \frac{1}{4}+\frac{1}{2}$
Hay $A< \frac{3}{4}$

27 tháng 3 2023

A = \(\dfrac{1}{1+2+3}\) + \(\dfrac{1}{1+2+3+4}\) +......+\(\dfrac{1}{1+2+3+4+....+59}\)

A = \(\dfrac{1}{(3+1).3:2}\) + \(\dfrac{1}{(4+1).4:2}\)+......+\(\dfrac{1}{(59+1).59:2}\)

A = \(\dfrac{2}{3.4}\) + \(\dfrac{2}{4.5}\) +.....+ \(\dfrac{2}{59.60}\)

A = 2.(\(\dfrac{1}{3.4}+\dfrac{1}{4.5}+....+\dfrac{1}{59.60}\))

A = 2. ( \(\dfrac{1}{3}\) - \(\dfrac{1}{4}\) + \(\dfrac{1}{4}\) - \(\dfrac{1}{5}\) +....+ \(\dfrac{1}{59}\) - \(\dfrac{1}{60}\))

A = 2. ( \(\dfrac{1}{3}\) - \(\dfrac{1}{60}\))

A = 2. \(\dfrac{19}{60}\)

A = \(\dfrac{19}{30}\)

5 tháng 5 2021

Dễ quá

17 tháng 2 2022

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