Find the odd one out in the following sequence: 1, 5, 14, 30, 50, 55, 91

Concept

A square pyramidal number is the running total of the squares of the first n natural numbers. The governing identity is:

12 + 22 + 32 + … + n2 = n(n+1)(2n+1)/6

Its first terms are 1, 5, 14, 30, 55, 91, 140, … . To spot an intruder, rebuild this exact series term by term and see which listed value breaks it.

Application

1. Term 1: 12 = 1

2. Term 2: add 22: 1 + 4 = 5

3. Term 3: add 32: 5 + 9 = 14

4. Term 4: add 42: 14 + 16 = 30

5. Term 5: add 52: 30 + 25 = 55

6. Term 6: add 62: 55 + 36 = 91

So the genuine series is 1, 5, 14, 30, 55, 91. The given list reads 1, 5, 14, 30, 50, 55, 91 — the values 55 and 91 already occupy the 5th and 6th positions correctly, while 50 sits where the pattern places nothing. The value 50 is the term that does not belong.

Cross-check

Verify with the closed form n(n+1)(2n+1)/6: for n = 5 it gives 5·6·11/6 = 55, and for n = 6 it gives 6·7·13/6 = 91 — both present in the list. No natural number n yields 50, confirming 50 as the odd one out.