As usual, we start by checking that the base case holds: if [tex]k=1[/tex] we have
[tex]1^2 = \dfrac{1\cdot 2 \cdot 3}{6}[/tex]
which is true. Now, we assume that
[tex]1^2+2^2+\ldots+k^2=\dfrac{k(k+1)(2k+1)}{6}[/tex]
and we check [tex]P(k+1)[/tex]: if we add the k+1-th term on both sides, we have
[tex]1^2+2^2+\ldots+k^2+(k+1)^2=\dfrac{k(k+1)(2k+1)}{6}+(k+1)^2[/tex]
And our goal is to rearrange the right hand term as follows:
[tex]\dfrac{k(k+1)(2k+1)}{6}+(k+1)^2 = \dfrac{2k^3+3k^2+k}{6}+\dfrac{6k^2+12k+6}{6}[/tex]
[tex]=\dfrac{2k^3+9k^2+13k+6}{6} = \dfrac{(2k+3)(k+1)(k+2)}{6}[/tex]
And we're done, because [tex](2k+3)(k+1)(k+2)[/tex] is exactly the formula [tex]k(k+1)(2k+1)[/tex] if you substitute k with k+1
