1
Prove that
for $n \in {\mathbb N}\text{.}$
The base case, $S(1): [1(1 + 1)(2(1) + 1)]/6 = 1 = 1^2$ is true. Assume that $S(k): 1^2 + 2^2 + \cdots + k^2 = [k(k + 1)(2k + 1)]/6$ is true. Then
and so $S(k + 1)$ is true. Thus, $S(n)$ is true for all positive integers $n\text{.}$