1. The problem is to prove the formula for the sum of the arithmetic sequence: $$1 + 5 + 9 + 13 + \dots + (4n - 3) = \frac{n(4n - 2)}{2}$$ 2. This sequence is arithmetic with the first term $a_1 = 1$ and common difference $d = 4$. 3. The $n$th term of an arithmetic sequence is given by $$a_n = a_1 + (n-1)d$$ Substituting, we get $$a_n = 1 + (n-1)4 = 4n - 3$$ which matches the given term. 4. The sum of the first $n$ terms of an arithmetic sequence is $$S_n = \frac{n}{2}(a_1 + a_n)$$ 5. Substitute $a_1 = 1$ and $a_n = 4n - 3$ into the sum formula: $$S_n = \frac{n}{2}(1 + 4n - 3) = \frac{n}{2}(4n - 2)$$ 6. Simplify the expression: $$S_n = \frac{n(4n - 2)}{2}$$ 7. This matches the formula given in the problem, so the formula is proven. Final answer: $$1 + 5 + 9 + 13 + \dots + (4n - 3) = \frac{n(4n - 2)}{2}$$