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Prove that 

$$1^2 + 2^2 + 3^2 + ... + n^2 = \dfrac{n (n + 1) (2n + 1)}{6}$$

For all positive integers n.
$$P(n) = 1^2 + 2^2 + 3^2 + ... + n^2 = \dfrac{n (n + 1) (2n + 1)}{6}$$
$$P(k) = \dfrac{k (k + 1) (2k + 1)}{6}$$
For $n = 1$
$$P(1) = \dfrac{1(1+1)(2(1)+1)}{6}$$
$$ = \dfrac{1(2)(3)}{6} = \dfrac{6}{6} = 1$$
$$P(n+1) = 1^2 + 2^2 + 3^2 + ... + n^2 + (n + 1)^2= \dfrac{n (n + 1) (2n + 1)}{ 6} + (n + 1)^2$$
For $(k+1)$
$$P(k+1) = a_k + (k + 1)^2$$
$$ = \dfrac{k (k + 1) (2k + 1)}{6} + (k + 1)^2$$
$$ = \dfrac{k (k + 1) (2k + 1) + 6(k + 1)^2}{6}$$
$$ = \dfrac{k(2k^2+3k+1) + 6(k^2+2k+1)}{6}$$
$$ = \dfrac{2k^3+3k^2+k + 6k^2+12k+6}{6}$$
$$ = \dfrac{2k^3+9k^2+13k+6}{6}$$
$$ = \dfrac{2k^3+9k^2+9k+4k+6}{6}$$
$$ = \dfrac{k(2k^2+9k+9)+4k+6}{6}$$
$$ = \dfrac{k(2k+3)(k+3)+2(2k+3)}{6}$$
$$ = \dfrac{(2k+3)[k(k+3)+2]}{6}$$
$$ = \dfrac{(2k+3)(k^2+3k+2)}{6}$$
$$ = \dfrac{(2k+3)(k+1)(k+2)}{6}$$
Let $w = k+1$
$$ = \dfrac{(w)(w+1)(2w+1)}{6}$$