In the number series 3, 14, 61, 254, 1033, the next term is _______. – UGC NET Question

This sequence is defined by a₁ = 3 and aₙ₊₁ = 4aₙ + n² + 1 for n ≥ 1. Applying the rule yields a₂ = 4·3 + 1² + 1 = 14, a₃ = 4·14 + 2² + 1 = 61, a₄ = 4·61 + 3² + 1 = 254 and a₅ = 4·254 + 4² + 1 = 1033, matching the given terms. For n = 5 the next term is a₆ = 4·1033 + 5² + 1 = 4158. Therefore 4158 is the unique number that continues the same recurrence correctly.

Option A:
Option A, 4126, is 32 less than the recurrence output for n = 5. To obtain 4126 we would need to subtract 32 from the formula result at the last step, which is not consistent with the earlier pattern. Hence this option does not follow the rule and is incorrect.

Option B:
Option B, 4142, is closer but still 16 less than the correct value 4158. It cannot be expressed as 4a₅ + 5² + 1 and would imply undercounting the contribution of the square term. As the recurrence is satisfied exactly before, option B cannot be the correct continuation.

Option C:
Option C, 4148, is slightly below the required value and also fails to satisfy the recurrence exactly. Selecting 4148 would again require modifying the output of the rule only at this step. Such a change destroys the strict structure of the sequence, so this option is not valid.

Option D:
Option D, 4158, equals exactly the result produced by aₙ₊₁ = 4aₙ + n² + 1 when n = 5. It maintains the process of quadrupling the previous term and adding a quadratic plus constant correction. Because this description explains each transition and leads naturally to 4158, option D is correct.