The sum $$1 + 3 + 11 + 25 + 45 + 71 + \ldots$$ upto 20 terms, is equal to

The given series is $$1,\,3,\,11,\,25,\,45,\,71,\ldots$$

First-order differences:
$$3-1 = 2,\; 11-3 = 8,\; 25-11 = 14,\; 45-25 = 20,\; 71-45 = 26,\ldots$$

Second-order differences:
$$8-2 = 6,\; 14-8 = 6,\; 20-14 = 6,\; 26-20 = 6,\ldots$$

The second difference is constant $$6$$, so the $$n^{\text{th}}$$ term $$a_n$$ is a quadratic expression:
$$a_n = An^{2}+Bn+C$$

For a quadratic sequence, the constant second difference equals $$2A$$, hence
$$2A = 6 \;\Longrightarrow\; A = 3$$

Thus $$a_n = 3n^{2}+Bn+C$$. Use the first three terms to find $$B,C$$.

Case 1: $$n = 1$$ gives $$3(1)^{2}+B(1)+C = 1 \;$$ ⇒ $$\;3 + B + C = 1$$ $$-(1)$$

Case 2: $$n = 2$$ gives $$3(2)^{2}+B(2)+C = 3 \;$$ ⇒ $$\;12 + 2B + C = 3$$ $$-(2)$$

Subtract $$(1)$$ from $$(2)$$:
$$(12 + 2B + C) - (3 + B + C) = 3 - 1$$
$$9 + B = 2 \;\Longrightarrow\; B = -7$$

Put $$B = -7$$ in $$(1)$$:
$$3 - 7 + C = 1 \;\Longrightarrow\; C = 5$$

Therefore the general term is
$$a_n = 3n^{2} - 7n + 5$$

We need the sum of the first $$20$$ terms:
$$S_{20} = \sum_{n=1}^{20} a_n = \sum_{n=1}^{20} \left(3n^{2} - 7n + 5\right)$$

Separate the sums:
$$S_{20} = 3\sum_{n=1}^{20} n^{2} \;-\; 7\sum_{n=1}^{20} n \;+\; 5\sum_{n=1}^{20} 1$$

Standard formulas:
$$\sum_{n=1}^{N} n = \frac{N(N+1)}{2}, \qquad \sum_{n=1}^{N} n^{2} = \frac{N(N+1)(2N+1)}{6}$$

For $$N = 20$$:
$$\sum n = \frac{20\cdot21}{2} = 210$$
$$\sum n^{2} = \frac{20\cdot21\cdot41}{6} = \frac{17220}{6} = 2870$$
$$\sum 1 = 20$$

Substitute these into $$S_{20}$$:
$$S_{20} = 3(2870) - 7(210) + 5(20)$$

Compute each term:
$$3(2870) = 8610$$
$$7(210) = 1470$$
$$5(20) = 100$$

Hence
$$S_{20} = 8610 - 1470 + 100 = 7240$$

Therefore, the required sum is $$7240$$.
Option A is correct.