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 \(S=\frac{4949}{9900}:2=\frac{4949}{9900}\cdot\frac{1}{2}=\frac{4949}{19800}\)

đặt 6 ra ngoài 

ta có \(\frac{1}{2}.6.\left(1+\frac{1}{4}+\frac{1}{10}+..............+\frac{1}{1540}\right)\)

=3 \(.\left(1+\frac{1}{1540}\right)\)

=3 \(.\frac{1541}{1540}\)

=3

=>3 > \(\frac{57}{462}\)

=> tích lớn hơn 

Là đặt \(\frac{1}{6}\) ra ngoài chứ bạn