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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 $$n \in \mathbb N.$$