Sum of Sequence of Squares/Proof by Induction

Proof
Proof by induction:

For all $n \in \N$, let $P \left({n}\right)$ be the proposition:
 * $\displaystyle \sum_{i \mathop = 1}^n i^2 = \frac {n \left({n + 1}\right)\left({2 n + 1}\right)} 6$

When $n = 0$, we see from the definition of vacuous sum that:
 * $0 = \displaystyle \sum_{i \mathop = 1}^0 i^2 = \frac {0 \left({1}\right) \left({1}\right)} 6 = 0$

and so $P(0)$ holds.

Base Case
When $n = 1$:
 * $\displaystyle \sum_{i \mathop = 1}^1 i^2 = 1^2 = 1$

Now, we have:
 * $\displaystyle \frac {n \left({n + 1}\right) \left({2 n + 1}\right)} 6 = \frac {1 \left({1 + 1}\right) \left({2 \cdot 1 + 1}\right)} 6 = \frac 6 6 = 1$

So $P(1)$ is true.

This is our base case.

Induction Hypothesis
Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 1$, then it logically follows that $P \left({k + 1}\right)$ is true.

So this is our induction hypothesis:
 * $\displaystyle \sum_{i \mathop = 1}^k i^2 = \frac {k \left({k + 1}\right) \left({2 k + 1}\right)} 6$

Then we need to show:
 * $\displaystyle \sum_{i \mathop = 1}^{k + 1} i^2 = \frac {\left({k + 1}\right) \left({k + 2}\right) \left({2 \left({k + 1}\right) + 1}\right)} 6$

Induction Step
This is our induction step:

Using the properties of summation, we have:
 * $\displaystyle \sum_{i \mathop = 1}^{k + 1} i^2 = \sum_{i \mathop = 1}^k i^2 + \left({k + 1}\right)^2$

We can now apply our induction hypothesis, obtaining:

So $P \left({k}\right) \implies P \left({k + 1}\right)$ and the result follows by the Principle of Mathematical Induction.

Therefore:
 * $\displaystyle \forall n \in \N: \sum_{i \mathop = 1}^n i^2 = \frac {n \left({n + 1}\right) \left({2 n + 1}\right)} 6$