Let $$a_n$$ be $$n^{th}$$ term of the series $$5 + 8 + 14 + 23 + 35 + 50 + \ldots$$ and $$S_n = \sum_{k=1}^{n} a_k$$. Then $$S_{30} - a_{40}$$ is equal to

Given: The series $$5 + 8 + 14 + 23 + 35 + 50 + \ldots$$

Finding the general term:

The first differences are: $$8-5=3, \; 14-8=6, \; 23-14=9, \; 35-23=12, \; 50-35=15, \ldots$$

The second differences are constant: $$6-3=3, \; 9-6=3, \; 12-9=3, \; 15-12=3$$

Since the second differences are constant (equal to 3), the first differences form an AP: $$3, 6, 9, 12, 15, \ldots$$ with the $$k$$-th difference being $$3k$$.

The $$n$$-th term is:

$$a_n = a_1 + \sum_{k=1}^{n-1}3k = 5 + 3 \cdot \frac{(n-1)n}{2} = \frac{3n^2 - 3n + 10}{2}$$

Verification: $$a_1 = \frac{3-3+10}{2} = 5$$ ✓, $$a_2 = \frac{12-6+10}{2} = 8$$ ✓, $$a_3 = \frac{27-9+10}{2} = 14$$ ✓

Finding $$a_{40}$$:

$$a_{40} = \frac{3(40)^2 - 3(40) + 10}{2} = \frac{4800 - 120 + 10}{2} = \frac{4690}{2} = 2345$$

Finding $$S_{30}$$:

$$S_{30} = \sum_{k=1}^{30}a_k = \sum_{k=1}^{30}\frac{3k^2 - 3k + 10}{2} = \frac{1}{2}\left(3\sum_{k=1}^{30}k^2 - 3\sum_{k=1}^{30}k + 10 \times 30\right)$$

Using summation formulas:

$$\sum_{k=1}^{30}k^2 = \frac{30 \times 31 \times 61}{6} = 9455$$

$$\sum_{k=1}^{30}k = \frac{30 \times 31}{2} = 465$$

Substituting:

$$S_{30} = \frac{1}{2}(3 \times 9455 - 3 \times 465 + 300) = \frac{1}{2}(28365 - 1395 + 300) = \frac{27270}{2} = 13635$$

Final answer:

$$S_{30} - a_{40} = 13635 - 2345 = 11290$$

The correct answer is Option C: 11290.