We need to evaluate the sum $\sum_{i=4}^{20} (2i - 5)$.

1. Problem Description

We need to evaluate the sum $\sum_{i=4}^{20} (2i - 5)$.

2. Solution Steps

We can split the summation into two separate summations:

$\sum_{i=4}^{20} (2i - 5) = \sum_{i=4}^{20} 2i - \sum_{i=4}^{20} 5$

$= 2 \sum_{i=4}^{20} i - \sum_{i=4}^{20} 5$

We can rewrite the first summation by using the formula for the sum of the first $n$ integers:

$\sum_{i=1}^{n} i = \frac{n(n+1)}{2}$

Therefore, $\sum_{i=4}^{20} i = \sum_{i=1}^{20} i - \sum_{i=1}^{3} i = \frac{20(20+1)}{2} - \frac{3(3+1)}{2} = \frac{20(21)}{2} - \frac{3(4)}{2} = \frac{420}{2} - \frac{12}{2} = 210 - 6 = 204$

The second summation is simply the sum of a constant, 5, from $i=4$ to $i=20$. There are $20-4+1 = 17$ terms in this summation. Thus,

$\sum_{i=4}^{20} 5 = 5 \times 17 = 85$

Substituting these values back into our original equation:

$2 \sum_{i=4}^{20} i - \sum_{i=4}^{20} 5 = 2(204) - 85 = 408 - 85 = 323$

3. Final Answer

323