\(=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{97.98}-\dfrac{1}{98.99}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{98.99}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{9702}\right)\)

\(=\dfrac{1}{2}.\dfrac{2425}{4851}\)

\(=\dfrac{2425}{9702}\)

=12(11.2−12.3+12.3−13.4+...+197.98−198.99)

=12(11.2−198.99)

=12(12−19702)

=12.24254851

=24259702

The last row will be :

\(A=\dfrac{2425}{4851}.\dfrac{1}{2}=\dfrac{2425}{9702}\)

Let \(A=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{97.98.99}\)

\(2A=\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+.....+\dfrac{2}{97.98.99}\)

\(2A=\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+....+\dfrac{1}{97.98}-\dfrac{1}{98.99}\)

\(2A=\dfrac{1}{1.2}-\dfrac{1}{98.99}=\dfrac{1}{2}-\dfrac{1}{98.99}=\dfrac{9702}{2.98.99}-\dfrac{2}{2.98.99}=\dfrac{9700}{19404}=\dfrac{2425}{4851}\)