Sum of Sequence of Squares/Proof by Induction

Proof
Proof by induction:

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

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

and so $P(0)$ holds.

Basis for the Induction
When $n = 1$:
 * $\displaystyle \sum_{i \mathop = 1}^1 i^2 = 1^2 = 1$

Now, we have:
 * $\displaystyle \frac {n \paren {n + 1} \paren {2 n + 1} } 6 = \frac {1 \paren {1 + 1} \paren {2 \times 1 + 1} } 6 = \frac 6 6 = 1$

and $P(1)$ is seen to hold.

This is our basis for the induction.

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

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

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

Induction Step
This is our induction step:

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

We can now apply our induction hypothesis, obtaining:

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

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