Answer
$6+\frac{31}{35} = \frac{241}{35}$
Work Step by Step
We sum from i = low bound to i = high bound.
$\frac{4(1)}{1+3} + \frac{4(2)}{2+3} + \frac{4(3)}{3+3} + \frac{4(4)}{4+3} = \frac{4}{4} + \frac{8}{5} + \frac{12}{6} + \frac{16}{7} = 1 + \frac{56}{35} + 2 + \frac{80}{35}$
$= 3+\frac{136}{35} = 6+\frac{31}{35} = \frac{241}{35}$