Làm lại câu a

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

\(2S=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{98.99}+\frac{1}{99.100}\)

\(2S=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\)

\(2S=1-\frac{1}{100}\)suy ra \(2S=\frac{99}{100}\)

\(S=\frac{99}{100}:2\)suy ra \(S=\frac{99}{200}\)

a, 2S=\(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{98.99.100}\)

\(2S=\frac{1}{1}-\frac{1}{2}+\frac{1}{3}+\frac{1}{2}-\frac{1}{3}+\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{100}\)

\(2S=1-\frac{1}{100}\)suy ra \(2S=\frac{99}{100}\)

\(S=\frac{99}{100}:2=\frac{99}{200}\)

Ta có 1/1.2.3 + 1/2.3.4 + 1/3.4.5 + ... + 1/98.99.100 

= 1/2 ( 1 / 1.2 - 1 / 2.3 + 1 / 2.3 - 1 / 3.4 + ...................+ 1 / 97.98 - 1 / 98.99 + 1 / 98.99 - 1 / 99.100) 

= 1 / 2 ( 1 / 1.2 - 1 / 99.100 ) 

= 4949 / 19800

Ta xét: \(\frac{1}{1.2}-\frac{1}{2.3}=\frac{2}{1.2.3};\frac{1}{2.3}-\frac{1}{3.4}=\frac{2}{2.3.4};...;\frac{1}{98.99}-\frac{1}{99.100}=\frac{2}{98.99.100}\)

Tổng quát : \(\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}=\frac{2}{n\left(n+1\right)\left(n+2\right)}\). Do đó:

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

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

\(=\frac{1}{1.2}-\frac{1}{99.100}=\frac{4949}{9900}\)

Vậy \(S=\frac{4949}{9900}\)

b, Ta có : \(\frac{1}{2}>\frac{57}{462}\)mà \(\frac{1}{6}+\frac{1}{24}+\frac{1}{60}+...+\frac{1}{9240}>0\)

nên A = \(\frac{1}{2}+\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{24}+\frac{1}{60}+...+\frac{1}{9240}\right)>\frac{57}{462}+0=\frac{57}{462}\)

Bạn sai Câu a.\(2S=\frac{4949}{9900}\)Vậy \(S=\frac{4949}{9900}:2=\frac{4949}{9900}\cdot\frac{1}{2}=\frac{4949}{19800}\)