1.13

The basis case is when $n=0$

$\sum _{i=1}^{0}i^{2}=0^{2}=0$

and using $n=0$ in the formula ${\frac {n(n+1)(2\cdot n+1)}{6}}$ you get:

${\frac {0(0+1)(2\cdot 0+1)}{6}}={\frac {0(1)(0+1)}{6}}={\frac {0}{6}}=0$

Since these are equal, the basis case is true.

Now, I will show that on the assumption that the summation is true for n, it follows that it is true for $n+1$

$\sum _{i=1}^{n+1}i^{2}=(n+1)^{2}+\sum _{i=1}^{n}i^{2}=(n^{2}+2n+1)+{\frac {n(n+1)(2\cdot n+1)}{6}}={\frac {6n^{2}+12n+6}{6}}+{\frac {2n^{3}+3n^{2}+n}{6}}={\frac {2n^{3}+9n^{2}+13n+6}{6}}$

Which should be equal to the formula ${\frac {n(n+1)(2\cdot n+1)}{6}}$ when $n=n+1$: ${\frac {(n+1)(n+1+1)(2\cdot (n+1)+1)}{6}}={\frac {(n+1)(n+2)(2n+2+1)}{6}}={\frac {(n^{2}+3n+2)(2n+3)}{6}}={\frac {2n^{3}+9n^{2}+13n+6}{6}}$

Back to Chapter 1.