Natural numbers are grouped into sets following a pattern: (1), (2, 3, 4), (5, 6, 7, 8, 9), continuing indefinitely.
Observe that the first set ends at \( 1^2 \), the second at \( 2^2 \), and the third at \( 3^2 \).
The 15th set terminates at \( 15^2 = 225 \). The preceding set, the 14th, concludes at \( 14^2 = 196 \). Consequently, the 15th set comprises integers from 197 through 225, inclusive.
The sum of the numbers in the 15th set is determined by the arithmetic series formula: \[ \text{Sum} = \frac{n}{2} \times (\text{First term} + \text{Last term}) \] Here: - The first term is 197. - The last term is 225. - The count of terms, \( n \), is \( 225 - 197 + 1 = 29 \). Plugging these values into the formula: \[ \text{Sum} = \frac{29}{2} \times (197 + 225) = \frac{29}{2} \times 422 = 6119 \]
The sum of the numbers in the 15th set is \( \boxed{6119} \).