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Prove that \(1^2 + 2^2 + 3^2 + \dots + n^2 = \tfrac{1}{6}(n(n+1)(2n+1))\) for all \(n \in \mathbb N.\)

Proof: We can see that \(1^2 = \tfrac{1}{6}(1\cdot 2 \cdot 3) \) and \(1^2 + 2^2 = \tfrac{1}{6}(2\cdot 3\cdot 5)\), so assume that \(1^2 + 2^2 + 3^2 + \dots + k^2 = \tfrac{1}{6}(k(k+1)(2k+1))\) for some \(k\in \mathbb N\). Continuing, we compute

\(1^2 + 2^2 + 3^2 + \dots + k^2 +(k+1)^2 \) = \(\tfrac{1}{6}(k(k+1)(2k+1)) +(k+1)^2 \)
= \(\tfrac{1}{6}(k+1)[k(2k+1) +6(k+1)]\)
= \(\tfrac{1}{6}(k+1)(2k^2 + 7k + 6)\)
= \(\tfrac{1}{6}(k+1)(k+2)(2k+3)\).

Thus \(1^2 + 2^2 + 3^2 + \dots + n^2 = \tfrac{1}{6}(n(n+1)(2n+1))\) for all \(n \in \mathbb N.\)

We can also begin with the induction assumption (after the basis step of course!) and continue with valid operations.

Proof: We can see that \(1^2 = \tfrac{1}{6}(1\cdot 2 \cdot 3) \) and \(1^2 + 2^2 = \tfrac{1}{6}(2\cdot 3\cdot 5)\), so assume that \(1^2 + 2^2 + 3^2 + \dots + k^2 = \tfrac{1}{6}(k(k+1)(2k+1))\) for some \(k\in \mathbb N\). Continuing, we compute

\(1^2 + 2^2 + 3^2 + \dots + k^2 = \tfrac{1}{6}(k(k+1)(2k+1)) \Rightarrow \)
\(1^2 + 2^2 + 3^2 + \dots + k^2+(k+1)^2 = \tfrac{1}{6}(k(k+1)(2k+1))+(k+1)^2 \Rightarrow \)
\(1^2 + 2^2 + 3^2 + \dots + k^2+(k+1)^2 = \tfrac{1}{6}(k+1)[k(2k+1) +6(k+1)] \Rightarrow \)
\(1^2 + 2^2 + 3^2 + \dots + k^2+(k+1)^2 = \tfrac{1}{6}(k+1)(2k^2 + 7k + 6)\Rightarrow \)
\(1^2 + 2^2 + 3^2 + \dots + k^2+(k+1)^2 = \tfrac{1}{6}(k+1)(k+2)(2k+3). \)

Thus \(1^2 + 2^2 + 3^2 + \dots + n^2 = \tfrac{1}{6}(n(n+1)(2n+1))\) for all \(n \in \mathbb N.\)