The problem asks to evaluate the summation $\sum_{i=4}^{20} (2i - 5)$.

1. Problem Description

The problem asks to evaluate the summation $\sum_{i=4}^{20} (2i - 5)$.

2. Solution Steps

We can split the summation into two separate summations:

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

We can factor out the constant 2 from the first summation:

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

We can calculate the summation of $i$ from 4 to 20 using the formula for the sum of the first $n$ integers: $\sum_{i=1}^{n} i = \frac{n(n+1)}{2}$.

Thus,

$\sum_{i=4}^{20} i = \sum_{i=1}^{20} i - \sum_{i=1}^{3} i = \frac{20(20+1)}{2} - \frac{3(3+1)}{2} = \frac{20(21)}{2} - \frac{3(4)}{2} = 10(21) - 3(2) = 210 - 6 = 204$.

So, $2 \sum_{i=4}^{20} i = 2(204) = 408$.

The second summation is simply summing the constant 5 a total of $20 - 4 + 1 = 17$ times:

$\sum_{i=4}^{20} 5 = 5(20 - 4 + 1) = 5(17) = 85$.

Therefore, the original summation is:

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

3. Final Answer

323