Answer
$550$
Work Step by Step
In order to set up the summation notation, we must calculate that value which keeps constantly changing. That value will be designated as the $i$ term. The remaining terms would remain constant in the summation notation.
Thus, we get the summation notation as:
$\sum_{i=1}^{20} (3i-4)=3 \sum_{i=1}^{20} i+(-4) \sum_{i=1}^{10} 1$
$=3 \sum_{i=1}^{20} i-4(20)$
$=3 \times \dfrac{20 \cdot 21}{2}+80$
$=550$
