The problem asks to evaluate the summation $\sum_{i=6}^{20} (2i - 5)$.

1. Problem Description

The problem asks to evaluate the summation $\sum_{i=6}^{20} (2i - 5)$.

2. Solution Steps

First, we can split the summation into two parts:

$\sum_{i=6}^{20} (2i - 5) = \sum_{i=6}^{20} 2i - \sum_{i=6}^{20} 5$

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

$\sum_{i=6}^{20} 2i = 2\sum_{i=6}^{20} i$

The number of terms in the summation is $20 - 6 + 1 = 15$.

Therefore, $\sum_{i=6}^{20} 5 = 5 \times 15 = 75$.

To evaluate $\sum_{i=6}^{20} i$, we can use the formula for the sum of the first $n$ integers:

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

We can rewrite the summation as follows:

$\sum_{i=6}^{20} i = \sum_{i=1}^{20} i - \sum_{i=1}^{5} i$

Using the formula:

$\sum_{i=1}^{20} i = \frac{20(20+1)}{2} = \frac{20 \times 21}{2} = 10 \times 21 = 210$

$\sum_{i=1}^{5} i = \frac{5(5+1)}{2} = \frac{5 \times 6}{2} = 5 \times 3 = 15$

So, $\sum_{i=6}^{20} i = 210 - 15 = 195$

Then, $2\sum_{i=6}^{20} i = 2 \times 195 = 390$

Finally, $\sum_{i=6}^{20} (2i - 5) = 390 - 75 = 315$

3. Final Answer

315