\(\left(1+\dfrac{1}{3}\right)\left(1+\dfrac{1}{8}\right)\left(1+\dfrac{1}{15}\right)...\left(1+\dfrac{1}{120}\right)\)
\(=\dfrac{4}{3}.\dfrac{9}{8}.\dfrac{16}{15}...\dfrac{121}{120}=\dfrac{2.2}{1.3}.\dfrac{3.3}{2.4}.\dfrac{4.4}{3.5}.\dfrac{11.11}{10.12}\)
\(=\dfrac{2}{1}.\dfrac{11}{12}=\dfrac{11}{6}\)