Naturally, a pyramid of zero layers doesn't need any snowballs, so P(0) = 0. Then using the given recurrence, we find
P(1) = 1
P(2) = 1 + 4 = 5
P(3) = 1 + 4 + 9 = 14
P(4) = 1 + 4 + 9 + 16 = 30
and so on; luckily, three of these are listed among the answer choices, which leaves 25 as an insufficient number of snowballs to make such a pyramid.
More generally, we would end up with
[tex]P(n) = 1^2 + 2^2 + 3^2 + \cdots + n^2 = \displaystyle \sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}6[/tex]
Then given some number of snowballs S, you could try to solve for n such that
S = n (n + 1) (2n + 1)/6
and any S that makes n a non-integer would be the answer.
