${\log }_3(1+\frac{1}{3})+{\log }_3(1+\frac{1}{4})+{\log }_3(1+\frac{1}{5})+\cdots +{\log }_3(1+\frac{1}{242})=$

$\begin{array}{l}{\log }_3(1+\frac{1}{3})+{\log }_3(1+\frac{1}{4})+{\log }_3(1+\frac{1}{5})+\cdots +{\log }_3(1+\frac{1}{242})\\ ={\log }_3\left(\frac{3+1}{3}\right)+{\log }_3\left(\frac{4+1}{4}\right)+\cdots +{\log }_3\left(\frac{242+1}{242}\right)\\ ={\log }_3\left(\frac{4}{3}\right)+{\log }_3\left(\frac{5}{4}\right)+\cdots +{\log }_3\frac{243}{242}\\ ={\log }_3\frac{4}{3}\times \frac{5}{4}\times \frac{6}{5}\times \cdots \times \frac{242}{241}\times \frac{243}{242}\\ \Rightarrow {\log }_3\frac{243}{3}={\log }_3{3}^4=4{\log }_{3}3=4\times 1=4\end{array}$