Identify the numbers that occur in the series 1, 7, 17, 31, 49, ... (A) 74 (B)…

Key insight: the differences between consecutive terms are 6, 10, 14, 18,... which increase by 4, so the sequence is quadratic with general term 2n^2 - 1.

• Verify the formula on given terms: for n = 1..5, 2n^2 - 1 gives 1, 7, 17, 31, 49, matching the series.

• Test each candidate number by solving 2n^2 - 1 = candidate:

• 74: 2n^2 - 1 = 74 ⇒ 2n^2 = 75 ⇒ n^2 = 37.5, not an integer ⇒ 74 is not in the series.

• 95: 2n^2 - 1 = 95 ⇒ 2n^2 = 96 ⇒ n^2 = 48, not a perfect square ⇒ 95 is not in the series.

• 97: 2n^2 - 1 = 97 ⇒ 2n^2 = 98 ⇒ n^2 = 49 ⇒ n = 7 ⇒ 97 is in the series.

• 127: 2n^2 - 1 = 127 ⇒ 2n^2 = 128 ⇒ n^2 = 64 ⇒ n = 8 ⇒ 127 is in the series.

• 161: 2n^2 - 1 = 161 ⇒ 2n^2 = 162 ⇒ n^2 = 81 ⇒ n = 9 ⇒ 161 is in the series.

Conclusion: The numbers that occur in the series from the given list are 97, 127 and 161.