Sum of Sequence of Squares/Proof by Induction

Theorem

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

Proof
Proof by induction:

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

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 $n=1$, we have $\displaystyle \sum_{i \mathop = 1}^1 i^2 = 1^2 = 1$.

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