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

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

\(A=2.\left(\frac{1}{1.2}-\frac{1}{99.100}\right)\)

\(A=2\cdot\frac{4949}{9900}=\frac{4949}{4950}\)

\(2A=\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{98.99.100}\)

\(2A=\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+\frac{5-3}{3.4.5}+...+\frac{100-98}{98.99.100}\)

\(2A=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{98.99}-\frac{1}{99.100}\)

\(2A=\frac{1}{2}-\frac{1}{99.100}=\frac{49}{99.100}\Rightarrow A=\frac{49}{2.99.100}\)

\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{98.99.100}\)

\(=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}+...+\frac{1}{98.99}-\frac{1}{99.100}\)

\(=\frac{1}{2}-\frac{1}{9900}=\frac{4950}{9900}-\frac{1}{9900}=\frac{4949}{9900}\)

Ta có:

\(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{98.99.100}\)

\(\Rightarrow\frac{1}{2}A=\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+....+\frac{2}{98.99.100}\)

\(\Rightarrow\frac{1}{2}A=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+....+\frac{1}{98.99}-\frac{1}{99.100}\)

\(\Rightarrow\frac{1}{2}A=\frac{1}{1.2}-\frac{1}{99.100}\)

\(\Rightarrow A=\frac{4949}{9900}.\frac{1}{2}=\frac{4949}{19800}\)

A= 1/1.2.3+1/2.3.4+1/3.4.5+...+1/98.99.100

A=1/1.2-1/2.3+1/2.3-1/3.4+1/3.4+...+1/98.99-1/99.100

A= 1/1.2 +(1/2.3-1/2.3)+(1/3.4-1/3.4)+...+(1/98.99+1/98.99)+1/99.100

A= 1/1.2 +0+0+0+...+0+1/99.100

A= 1/2+1/9900

A=4950/9900+ 1/9900

A= 4951/9900

\(2S=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{99.100}\)

\(2S=\frac{1}{2}-\frac{1}{9900}\)

\(2S=\frac{4949}{9900}\)

\(S=\frac{4949}{19800}\)

Ta xét : \(\frac{1}{1.2}-\frac{1}{2.3}=\frac{2}{1.2.3}\)

\(\frac{1}{2.3}-\frac{1}{3.4}=\frac{2}{2.3.4}\)

...

\(\frac{1}{98.99}-\frac{1}{99.100}=\frac{2}{98.99.100}\)

Ta có : 2S = \(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{98.99}-\frac{1}{99.100}\)

=> 2S = \(\frac{1}{1.2}-\frac{1}{99.100}\)

=> 2S = \(\frac{4949}{9900}\)

=> S = \(\frac{4949}{19800}\)

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

2S= \(1-\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)2S= 1- \(\dfrac{1}{100}\)

2S= \(\dfrac{99}{100}\)

S= \(\dfrac{99}{100}.\dfrac{1}{2}\)

S=\(\dfrac{198}{100}\)

Coi \(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{98.99.100}\)

\(\Rightarrow2A=2x\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{98.99.100}\right)\)

\(=\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{98.99.100}\)

\(=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{98.99}-\frac{1}{99.100}\)

\(=\frac{1}{1.2}-\frac{1}{99.100}\)

\(=\frac{4950}{9900}-\frac{1}{9900}\)

\(=\frac{4949}{9900}\)

các bn làm đúng rồi

tk mk nha

thnak

Gọi tổng trên là A

A=1/1.2.3+1/2.3.4+1/3.4.5+...1/98.99.100

Ta xét :

1/1.2 - 1/2.3 = 2/1.2.3; 1/2.3 - 1/3.4 = 2/2.3.4;...; 1/98.99 - 1/99.100 = 2/98.99.100  

tổng quát: 1/n(n+1) - 1/(n+1)(n+2) = 2/n(n+1)(n+2).

Do đó:  2A = 2/1.2.3 + 2/2.3.4 + 2/3.4.5 +...+ 2/98.99.100

 = (1/1.2 - 1/2.3) + (1/2.3 - 1/3.4) +...+ (1/98.99 - 1/99.100)  

= 1/1.2 - 1/2.3 + 1/2.3 - 1/3.4 + ... + 1/98.99 - 1/99.100  

= 1/1.2 - 1/99.100  

= 1/2 - 1/9900  

= 4950/9900 - 1/9900  

= 4949/9900.  

Vậy A = 4949 / 9900