The problem is to calculate the sum $\sum_{i=5}^{20} (2i + 5)$.

1. Problem Description

The problem is to calculate the sum $\sum_{i=5}^{20} (2i + 5)$.

2. Solution Steps

We can split the summation into two parts:

$\sum_{i=5}^{20} (2i + 5) = \sum_{i=5}^{20} 2i + \sum_{i=5}^{20} 5$.

The first summation can be written as $2 \sum_{i=5}^{20} i$.

The second summation is simply adding the constant 5 a total of $20 - 5 + 1 = 16$ times, so it equals $5 \times 16 = 80$.

We can calculate $\sum_{i=5}^{20} i$ using the formula for the sum of the first n integers:

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

So, $\sum_{i=5}^{20} i = \sum_{i=1}^{20} i - \sum_{i=1}^{4} i = \frac{20(20+1)}{2} - \frac{4(4+1)}{2} = \frac{20 \times 21}{2} - \frac{4 \times 5}{2} = 10 \times 21 - 2 \times 5 = 210 - 10 = 200$.

Therefore, $2 \sum_{i=5}^{20} i = 2 \times 200 = 400$.

Finally, $\sum_{i=5}^{20} (2i + 5) = 400 + 80 = 480$.

3. Final Answer

480