\(\dfrac{1}{1\cdot2}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{49\cdot50}=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}\\ =\left(1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{49}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{50}\right)\\ =\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{50}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{50}\right)\\ =1+\dfrac{1}{2}+...+\dfrac{1}{50}-1-\dfrac{1}{2}-\dfrac{1}{3}-...-\dfrac{1}{25}\\ =\dfrac{1}{26}+\dfrac{1}{27}+...+\dfrac{1}{50}\)

Ta có:
1/1.2 + 1/3.4 + 1/5.6 + ... + 1/49.50 = 1/26 + 1/27 + 1/28 + .. + 1/50
Xét vế trái:
1/1.2 + 1/3.4 + 1/5.6 + ... + 1/49.50
= 1 - 1/2 + 1/3 - 1/4 + 1/5 - 1/6 + ... + 1/49 - 1/50
= ( 1 + 1/3 + 1/5 + ... + 1/49 ) - ( 1/2 + 1/4 + 1/6 + ... + 1/50 )
= ( 1 + 1/3 + 1/5 + ... + 1/49 ) + (1/2 + 1/4 + 1/6 + ... + 1/50 ) - 2 . ( 1/2 + 1/4 + 1/6 + ... + 1/50 )
= ( 1 + 1/2 + 1/3 + 1/4 + ...+ 1/49 + 1/50 ) - ( 1 + 1/2 + 1/3 + ... + 1/25 )
= 1/26 + 1/27 + 1/28 + ... + 1/49 + 1/50 (1)
Từ (1) => Vế trái = Vế phải 
=> Điều phải chứng minh 
- HokTot - 

\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{49.50}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\)

\(=1-\dfrac{1}{50}< 1\)

Lời giải:

\(\text{VT}=\frac{1}{1.2}+\frac{1}{3.4}+....+\frac{1}{49.50}\)

\(=\frac{2-1}{1.2}+\frac{4-3}{3.4}+....+\frac{50-49}{49.50}\)

\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)

\(=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{49}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}....+\frac{1}{50}\right)\)

\(=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{49}+\frac{1}{50}-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}....+\frac{1}{50}\right)\)

\(=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{49}+\frac{1}{50}-\left(1+\frac{1}{2}+\frac{1}{3}....+\frac{1}{25}\right)\)

\(=\frac{1}{26}+\frac{1}{27}+....+\frac{1}{49}+\frac{1}{50}\)

Do đó ta có đpcm.

\(A=\dfrac{1}{1\cdot2\cdot3}+\dfrac{1}{2\cdot3\cdot4}+...+\dfrac{1}{2021\cdot2022\cdot2023}\)
\(=\dfrac{1}{2}\left(\dfrac{2}{1\cdot2\cdot3}+\dfrac{2}{2\cdot3\cdot4}+...+\dfrac{2}{2021\cdot2022\cdot2023}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{1}{1\cdot2}-\dfrac{1}{2\cdot3}+\dfrac{1}{2\cdot3}-\dfrac{1}{3\cdot4}+...+\dfrac{1}{2021\cdot2022}-\dfrac{1}{2022\cdot2023}\right)\)

\(=\dfrac{1}{2}\cdot\left(\dfrac{1}{2}-\dfrac{1}{4090506}\right)\)

\(=\dfrac{1}{2}\cdot\dfrac{2045252}{4090506}=\dfrac{1022626}{4090506}=\dfrac{511313}{2045253}\)

`1/(1.2.3) + 1/(2.3.4) + ... + 1/(2021 . 2022 .2023)`

`=> 2/(1.2.3) + 2/(2.3.4) + ... + 2/(2021 . 2022. 2023)`

`= 1/(1.2) - 1/(2.3) + 1/(2.3) - 1/(3.4) + ... + 1/(2021.2022) - 1/(2022 . 2023)`

`= 1/2 - 1/4090506`

`=4090506/8181012 - 2/8181012`

`= 4090504/8181012`

A=\(\dfrac{1}{1\cdot2\cdot3}+\dfrac{1}{2\cdot3\cdot4}+...+\dfrac{1}{2014\cdot2015\cdot2016}=\dfrac{1}{2}\cdot\left(\dfrac{1}{1\cdot2}-\dfrac{1}{2\cdot3}+\dfrac{1}{2\cdot3}-\dfrac{1}{3\cdot4}+...+\dfrac{1}{2014\cdot2015}-\dfrac{1}{2015\cdot2016}\right)=\dfrac{1}{2}\cdot\left(\dfrac{1}{2}-\dfrac{1}{2015}\cdot\dfrac{1}{2016}\right)=\dfrac{1}{4}-\dfrac{1}{2\cdot2015\cdot2016}< \dfrac{1}{4}\)

Vậy A<\(\dfrac{1}{4}\)

---bé hơn hoặc bằng tức là chỉ cần xảy ra 1 khả năng cũng thõa mãn nhé---

mình

bài này tương tự bài trên

\(S=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+...+\dfrac{1}{99.100.101}\)

\(S=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+\dfrac{1}{3.4}-\dfrac{1}{4.5}+...+\dfrac{1}{99.100}-\dfrac{1}{100.101}\right)\)

\(S=\dfrac{1}{4}-\dfrac{1}{2.100.101}\)

\(B=\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+\dfrac{2}{3.4.5}+\dfrac{2}{4.5.6}+\dfrac{2}{5.6.7}+\dfrac{2}{6.7.8}\)

\(=\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{6.7}-\dfrac{1}{7.8}\)

\(=\dfrac{1}{1.2}-\dfrac{1}{7.8}\)

\(=\dfrac{1}{2}-\dfrac{1}{56}=\dfrac{27}{56}\)