The $nth$ term of the series $4, 14, 30, 52, 80, 114, ...$ is
- $n^{2} + n + 2$
- $3n^{2} + n$
- $3n^{2} - 5n + 2$
- $(n + 1)^{2}$
Series: 4, 14, 30, 52, 80, 114. Differences: 10, 16, 22, 28, 34. Second differences: 6, 6, 6, 6. This is a quadratic sequence an^2 + bn + c. 2a = 6 => a = 3. 3a + b = 10 => 9 + b = 10 => b = 1. a + b + c = 4 => 3 + 1 + c = 4 => c = 0. Formula: 3n^2 + n.
Examine the differences between the terms to find the underlying pattern. The differences are 10, 16, 22, 28, and 34, which form an arithmetic progression with a common difference of 6. Because the second differences are constant, the nth term is a quadratic polynomial of the form an^2 + bn + c. Using the first three terms to form the equations a + b + c = 4, 4a + 2b + c = 14, and 9a + 3b + c = 30, solving the system yields a = 3 and b = 1 and c = 0. Therefore, the nth term is 3n^2 + n.