TADM2E 1.11
The basis case is when <math>n = 0</math><br>
- <math>\sum_{i=1}^0 i^2 = 0^2 = 0 </math><br>
and using <math>n=0</math> in the formula <math>\frac {n(n + 1)(2 \cdot n + 1)} {6}</math> you get:<br>
- <math>\frac {0(0 + 1)(2 \cdot 0 + 1)} {6} = \frac {0(1)(0 + 1)} {6} = \frac {0} {6} = 0</math><br>
Since these are equal, the basis case is true.<br>
Now, I will show that on the assumption that the summation is true for n, it follows that it is true for <math>n+1</math>
- <math>\sum_{i = 1}^{n + 1} i^2 =
(n + 1)^2 + \sum_{i = 1}^n i^2 = (n^2 + 2n + 1) + \frac{n(n + 1)(2 \cdot n + 1)} {6} = \frac{6n^2 + 12n + 6}{6}+ \frac{2n^3 + 3n^2 + n} {6} = \frac {2n^3 + 9n^2 + 13n + 6} {6}</math><br>
Which should be equal to the formula <math>\frac {n(n + 1)(2 \cdot n + 1)} {6}</math> when <math>n = n + 1</math>: <math>\frac {(n + 1)(n + 1 + 1)(2 \cdot (n + 1) + 1)} {6}= \frac {(n + 1)(n + 2)(2n + 2 + 1)} {6}= \frac {(n^2 + 3n + 2)(2n + 3)} {6}= \frac {2n^3 + 9n^2 + 13n + 6} {6}</math>
