Prove using induction that, for all \( n \in \mathbb{N} \), the sum of the first \( n \) square numbers can be found using the formula: \[ 1^{2}+2^{2}+3^{2}+4^{2}+\cdots+n^{2}=\frac{n(n+1)(2 n+1)}{6} \]

To prove this statement using mathematical induction, we will follow two main steps: the base case and the inductive step. **Base Case:** First, let's verify the formula for \( n = 1 \): \[ 1^2 = 1 \] According to the formula, we have: \[ \frac{1(1+1)(2 \cdot 1 + 1)}{6} = \frac{1 \cdot 2 \cdot 3}{6} = \frac{6}{6} = 1 \] The base case holds true since both sides equal 1. **Inductive Step:** Now we assume that the formula holds for some arbitrary positive integer \( k \); that is, we assume: \[ 1^{2}+2^{2}+3^{2}+\cdots+k^{2} = \frac{k(k+1)(2k+1)}{6} \] We need to prove that it holds for \( k + 1 \): \[ 1^{2}+2^{2}+3^{2}+\cdots+k^{2} + (k + 1)^{2} = \frac{(k+1)(k+2)(2(k+1)+1)}{6} \] Using the inductive hypothesis, we can rewrite the left-hand side: \[ \frac{k(k+1)(2k+1)}{6} + (k + 1)^{2} \] Factoring out \( (k + 1) \): \[ = \frac{(k + 1)}{6} (k(2k + 1) + 6(k + 1)) \] Now simplify the expression inside the parentheses: \[ k(2k + 1) + 6(k + 1) = 2k^2 + k + 6k + 6 = 2k^2 + 7k + 6 \] Next, we can factor the quadratic: \[ 2k^2 + 7k + 6 = (2k + 3)(k + 2) \] Therefore, we can write: \[ = \frac{(k + 1)(2k + 3)(k + 2)}{6} \] Now let's rewrite the right-hand side for \( k + 1 \): \[ \frac{(k + 1)(k + 2)(2(k + 1) + 1)}{6} = \frac{(k + 1)(k + 2)(2k + 2 + 1)}{6} = \frac{(k + 1)(k + 2)(2k + 3)}{6} \] Both sides are now equal. Therefore, we have proved that: \[ 1^{2}+2^{2}+3^{2}+\cdots+n^{2} = \frac{n(n+1)(2 n+1)}{6} \] holds true for all \( n \in \mathbb{N} \) by induction.