Answer
$\displaystyle\sum_{k=1}^{n}\frac{3^{k}}{k}$
Work Step by Step
There are $n$ terms ( terms contain from $(3)^{1}$ to $(3)^{n})$.
Count them with index k: $\displaystyle \sum_{k=1}^{n}$(...)
The first term (when k=$1$) is $3=\displaystyle \frac{3^{1}}{1}$,
The second term (when k=$2$) is $\displaystyle \frac{3^{2}}{2}$,
The third term (when k=$3$) is $\displaystyle \frac{3^{3}}{3}$,
The kth is $\displaystyle \frac{3^{k}}{k}$ so
$3+\displaystyle \frac{3^{2}}{2}+\frac{3^{3}}{3}+\dots+\frac{3^{n}}{n}=\sum_{k=1}^{n}\frac{3^{k}}{k}$
