Sum of the first n squares

Example — Sum of the first $n$ squares

$P(n)\colon\quad 1^2+2^2+\cdots+n^2 = \frac{n(n+1)(2n+1)}{6}.$ Base: $n=1$: $1=\dfrac{1\cdot 2\cdot 3}{6}$. True.

Step: we assume $P(k)$ and add $(k+1)^2$: $\frac{k(k+1)(2k+1)}{6}+(k+1)^2 = \frac{(k+1)\bigl[k(2k+1)+6(k+1)\bigr]}{6} = \frac{(k+1)(2k^2+7k+6)}{6} = \frac{(k+1)(k+2)(2k+3)}{6},$ which is $P(k+1)$. ∎

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Topics: Set theory
Concepts: Principle of induction
Methods: Induction
Skills: Proving