The sum of the first 20 terms of the series $$5 + 11 + 19 + 29 + 41 + \ldots$$ is

The series is: 5, 11, 19, 29, 41, ...

The differences between consecutive terms are: 6, 8, 10, 12, ... (an arithmetic progression with common difference 2).

The general term can be found as:

$$ a_n = n^2 + 3n + 1 $$

Verification: $$a_1 = 1+3+1 = 5$$ ✓, $$a_2 = 4+6+1 = 11$$ ✓, $$a_3 = 9+9+1 = 19$$ ✓

The sum of first 20 terms is:

$$ S_{20} = \sum_{n=1}^{20}(n^2 + 3n + 1) = \sum_{n=1}^{20} n^2 + 3\sum_{n=1}^{20} n + 20 $$

Using standard formulas:

$$ \sum_{n=1}^{20} n^2 = \frac{20 \times 21 \times 41}{6} = 2870 $$

$$ \sum_{n=1}^{20} n = \frac{20 \times 21}{2} = 210 $$

$$ S_{20} = 2870 + 3(210) + 20 = 2870 + 630 + 20 = 3520 $$

The correct answer is 3520.