* Chứng tỏ

Ta có :\(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+...+\dfrac{1}{98.99.100}\)

= \(\dfrac{1}{1.2.3}.\dfrac{2}{2}+\dfrac{1}{2.3.4}.\dfrac{2}{2}+...+\dfrac{1}{98.99.100}.\dfrac{2}{2}\)

= \(\dfrac{1}{2}.\left(\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+\dfrac{2}{3.4.5}+...+\dfrac{2}{98.99.100}\right)\)

= \(\dfrac{1}{2}.\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{98.99}-\dfrac{1}{99.100}\right)\)

= \(\dfrac{1}{2}.\left(\dfrac{1}{1.2}+0+0+...+0+\dfrac{-1}{99.100}\right)\)

= \(\dfrac{1}{2}.\left(\dfrac{1}{2}+\dfrac{-1}{9900}\right)\)

= \(\dfrac{1}{2}.\left(\dfrac{4850}{9900}+\dfrac{-1}{9900}\right)\)

= \(\dfrac{1}{2}.\dfrac{4849}{9900}\)

= \(\dfrac{4849}{19800}\)

* So sánh

\(\dfrac{4950}{19800}\) và \(\dfrac{1}{4}\)

\(\dfrac{1}{4}=\dfrac{4950}{19800}\)

Vì \(\dfrac{4950}{19800}=\dfrac{4950}{19800}\)

=> Tổng trên bằng với\(\dfrac{1}{4}\)

mình ko chắc là đúng

\(2A=\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+\dfrac{2}{3.4.5}+...+\dfrac{2}{18.19.20}\)

\(\Rightarrow2A=\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+\dfrac{1}{3.4}-\dfrac{1}{4.5}+...+\dfrac{1}{18.19}-\dfrac{1}{19.20}\)

\(\Rightarrow2A=\dfrac{1}{1.2}-\dfrac{1}{19.20}< \dfrac{1}{1.2}\)

\(\Rightarrow2A< \dfrac{1}{2}\)

\(\Rightarrow A< \dfrac{1}{4}\) (đpcm)

lỗi

lỗi cực kỳ

a, A= 1/2. (2/1.2.3+2/2.3.4+2/3.4.5+...+2/18.19.20) A=1/2. (1/1.2-1/2.3+1/2.3-1/3.4+1/3.4-1/4.5+...+1/18.19-1/19.20) A=1/2. (1/1.2-1/19.20) A=1/2. 189/380 A= 189/760

\(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{10.11.12}\)

\(=\dfrac{1}{2}.\left(\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+...+\dfrac{2}{10.11.12}\right)\)

\(=\dfrac{1}{2}.\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{10.11}-\dfrac{1}{11.12}\right)\)

\(=\dfrac{1}{2}.\left(\dfrac{1}{1.2}-\dfrac{1}{11.12}\right)\)

\(=\dfrac{1}{2}.\left(\dfrac{1}{2}-\dfrac{1}{132}\right)\)

\(=\dfrac{1}{2}.\dfrac{65}{132}=\dfrac{65}{264}\)

Tìm y:

-y:1/2-5/2=4+1/2

-y:1/2 = 4+1/2+5/2

-y:1/2 = 7

-y = 7.2

y = -14

Vậy y = -14

\(A=\dfrac{3}{4}\cdot\dfrac{8}{9}\cdot\dfrac{15}{16}\cdot...\cdot\dfrac{899}{900}\)

\(A=\dfrac{1\cdot3}{2\cdot2}\cdot\dfrac{2\cdot4}{3\cdot3}\cdot\dfrac{3\cdot5}{4\cdot4}\cdot...\cdot\dfrac{29\cdot31}{30\cdot30}\)

\(A=\dfrac{1\cdot\left(2\cdot3\cdot4\cdot5\cdot...\cdot29\right)^2\cdot30\cdot31}{\left(2\cdot3\cdot4\cdot...\cdot30\right)^2}\)

\(A=\dfrac{1\cdot\left(2\cdot3\cdot4\cdot5\cdot...\cdot29\right)^2\cdot30\cdot31}{\left(2\cdot3\cdot4\cdot5\cdot...\cdot29\right)^2\cdot30\cdot30}\)

\(A=\dfrac{1\cdot31}{30}=\dfrac{31}{30}\)

Ta có : \(\dfrac{1}{101}>\dfrac{1}{300}\)

...

\(\dfrac{1}{299}>\dfrac{1}{300}\)

Do đó :

\(\dfrac{1}{101}+\dfrac{1}{102}+..+\dfrac{1}{300}>\dfrac{1}{300}+\dfrac{1}{300}..+\dfrac{1}{300}\)

\(\Rightarrow\dfrac{1}{101}+\dfrac{1}{102}+..+\dfrac{1}{300}>\dfrac{200}{300}=\dfrac{2}{3}\)

Vậy...

A=3/2².8/3².15/4²......899/30² A=3.8.15.....899/2².3².4².....30² A=(1.3).(2.4).(3.5).....(29.31)/(2.3.4....30)(2.3.4...30) A=(1.2.3....29).(3.4.5...31)/(2.3.4...30)(2.3.4...30) A=1.31/30.2=31/60

câu b bài 2:

\(\dfrac{1^2}{1\cdot2}\cdot\dfrac{2^2}{2\cdot3}\cdot\dfrac{3^2}{3\cdot4}\cdot\dfrac{4^2}{4\cdot5}\)

\(=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot\dfrac{4}{5}\)

\(=\dfrac{1}{5}\)

câu a bài 2:

\(\dfrac{1}{1\cdot2\cdot3}+\dfrac{1}{2\cdot3\cdot4}+\dfrac{1}{3\cdot4\cdot5}+...+\dfrac{1}{10\cdot11\cdot12}\)

\(=\dfrac{1}{1}-\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{4}-...-\dfrac{1}{12}\)

\(=1-\dfrac{1}{12}=\dfrac{11}{12}\)

a) Ta có:

3A= \(1+\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}\left(1\right)\)

A= \(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{100}}\left(2\right)\)

Lấy (1) - (2) ta được:

1-\(\dfrac{1}{3^{100}}\)

b) Ta xét:

\(\dfrac{1}{1.2}-\dfrac{1}{2.3}=\dfrac{2}{1.2.3},...,\dfrac{1}{37.38}-\dfrac{1}{38.39}=\dfrac{2}{37.38.39}\)

Ta có:

2B=\(\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+..+\dfrac{2}{37.38.39}\)

=\(\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}\right)+\left(\dfrac{1}{2.3}-\dfrac{1}{3.4}\right)+..+\left(\dfrac{1}{37.38}-\dfrac{1}{38.39}\right)\)

=\(\dfrac{1}{1.2}-\dfrac{1}{38.39}=\dfrac{740}{38.39}=\dfrac{370}{741}\)

Vậy \(\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+\dfrac{2}{3.4.5}+..+\dfrac{2}{37.38.39}\)

=\(\dfrac{370}{741}\)

Nếu bn cảm thấy mk đúng tick cho mk nhé!