Answer
$$\frac{{73}}{{12}}$$
Work Step by Step
$$\eqalign{
& \sum\limits_{i = 1}^4 {\frac{{i + 1}}{i}} \cr
& {\text{Evaluate }}i = 1{\text{ to }}i = 4 \cr
& i = 1 \cr
& = \frac{{1 + 1}}{1} = 2 \cr
& i = 2 \cr
& = \frac{{2 + 1}}{2} = \frac{3}{2} \cr
& i = 3 \cr
& = \frac{{3 + 1}}{3} = \frac{4}{3} \cr
& i = 4 \cr
& = \frac{{4 + 1}}{4} = \frac{5}{4} \cr
& \cr
& {\text{The sum is}} \cr
& \sum\limits_{i = 1}^5 {\left( {{i^2} + i} \right)} = 2 + \frac{3}{2} + \frac{4}{3} + \frac{5}{4} \cr
& \sum\limits_{i = 1}^5 {\left( {{i^2} + i} \right)} = \frac{{73}}{{12}} \cr} $$
