Sum of Sequence of Squares

Theorem

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

Proof
Proof by induction:

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

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

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:
 * $\displaystyle \sum_{i=1}^k i^2 = \frac {k \left({k + 1}\right) \left({2 k + 1}\right)} 6$

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

Induction Step
This is our induction step:

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

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

Proof by Telescoping Sum
Observe that $3 i \left({i + 1}\right) = i \left({i + 1}\right) \left({i + 2}\right) - i \left({i + 1}\right) \left({i - 1}\right)$.

By taking the sum we'll get a telescoping one on the RHS and the conclusion follows.

Historical Note
This result was documented by Āryabhaṭa in his work Āryabhaṭīya of 499 CE.