Answer
$\dfrac{73}{12} \quad \text{or}\quad 6\frac{1}{12}$
Work Step by Step
After substituting $n=1,2,3,4$, we can have the $4$ terms of the given sequence.
\begin{align*}
a_1&=2\\
\\a_2&=\dfrac{2+1}{2}=\dfrac{3}{2}\\
\\a_3&=\dfrac{3+1}{3}=\dfrac{4}{3}\\
\\a_4&=\dfrac{4+1}{4}=\dfrac{5}{4}
\end{align*}
Thus, the sum of the four terms is
$S_4= 2+\dfrac{3}{2}+\dfrac{4}{3}+\dfrac{5}{4}=\dfrac{73}{12}=6\frac{1}{12}$
