Sum of Sequence of Squares

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 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.

Here are the details: First, note that

where $T_n$ is the $n$th Triangular number.

Now the telescoping sum above comes into play. Note that $i\left(i+1\right)$ must be even, which means that $3i\left(i+1\right)$ is just 6 times the $i$th Triangular number: $6\left(\dfrac{i\left(i+1\right)}{2}\right)$.

Here are the first few terms in the sequence $3i\left(i+1\right)$, after dividing each side by 6:

$T_1=1=\dfrac{1\cdot2\cdot3}{6}$

$T_2=3=\dfrac{2\cdot3\cdot4}{6}-\dfrac{1\cdot2\cdot3}{6}$

$T_3=6=\dfrac{3\cdot4\cdot5}{6}-\dfrac{2\cdot3\cdot4}{6}$

$T_4=10=\dfrac{4\cdot5\cdot6}{6}-\dfrac{3\cdot4\cdot5}{6}$

If we continue on and sum the first $n$ terms, we see that

$\displaystyle \sum_{i=1}^nT_i=\dfrac{n\left(n+1\right)\left(n+2\right)}{6}$.

Now back to the original sum in question:

Direct Proof

 * [[File:Sum of Sequences of Squares.jpg]]

We can observe from the above diagram that:
 * $\displaystyle \forall n \in \N: \sum_{i=1}^n i^2 = \sum_{i=1}^n \left({\sum_{j=i}^n j}\right)$

Therefore we have:

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