Proof Problem - 4

First, let us see if this is true for $\large n = 1$ ,

$\large \frac {1(1 + 1)(2 + 1)}{6} = \frac {1(2)(3)}{6} = \frac {6}{6} = 1$ , $\large 1^2 = 1$ so it is true for $\large n = 1$ ,

Let us assume that it is true for $\large n = r$ , so that $\large 1^2 + 2^2 + 3^2 + \dots + r^2 = \frac {r(r + 1)(2r + 1)}{6}$ ,

Then $\large 1^2 + 2^2 + 3^2 + \dots + r^2 + (r + 1)^2 = \frac {r(r + 1)(2r + 1)}{6} + (r + 1)^2$ ,

That equals $\large \frac {r(r + 1)(2r + 1) + 6(r + 1)^2}{6} = \frac {(r + 1)[r(2r + 1) + 6(r + 1)]}{6}$ ,

Which equals $\large \frac {(r + 1)({2r}^2 + r)(6r + 6)}{6} = \frac {(r + 1)({2r}^2 + 7r + 6)}{6}$ ,

Finally, that equals $\large \frac {(r + 1)(r + 2)(2r + 3)}{6} = \frac {(r + 1)(r + 2)(2r + 2) + 1)}{6} = \frac {(r + 1)(r + 2)(2(r + 1) + 1)}{6}$ .

If this is true for $\large r$ and $\large r + 1$ , it is true for all positive integers. Q.E.D.