Video: Sum of Consecutive Squares

[Level 0]
Some things really are easier shown than said. Here's an uber-concise graphical 'proof' of the famous identity
\[
\sum_{i=1}^{n}i^{2}=\dfrac{n\left(n+1\right)\left(2n+1\right)}{6}
\]
(For those unfamiliar with \(\Sigma\) notation, here's a quick explanation. Trust me, it's not as scary as it looks!)




Most of the demonstrations I've seen on YouTube last about 11-12 minutes, which I think is far too long to graphically explicate this simple identity. A fifty-one second clip should be sufficient, provided you watch it a few times.

In the unlikely case that several viewings still leave you befuddled, here are the thousand words the video aims to paint:


The "Thousand Words"

Essentially what we're doing here is this. We recall that $n^{2}=1+3+\cdots+\left(2n-1\right)$. So when we sum $1+\cdots+n^{2}$, we have $n$ copies of $1$, $\left(n-1\right)$ copies of $3$, and so on, until we have one copy of $2n-1$. So we have
\begin{eqnarray*}
\sum_{i=1}^{n}i^{2} & = & \sum_{i=1}^{n}\sum_{j=1}^{i}\left(2j-1\right)\\
 & = & \sum_{j=1}^{n}\left(n-j+1\right)\left(2j-1\right)\\
 & = & \sum_{k=1}^{n}\left(k\right)\left(2n-2k+1\right)
\end{eqnarray*}
This is the yellow triangular wedge.

Next, we write the sum $1+\cdots+n^{2}$ in a more traditional way:
$1=1$, $2^{2}=2+2$, $3^{2}=3+3+3$, and so on. So we have
\[
\sum_{i=1}^{n}i^{2}=\sum_{k=1}^{n}k\left(k\right)
\]
Combining these two, we have
\begin{eqnarray*}
\sum_{i=1}^{n}i^{2}+\sum_{i=1}^{n}i^{2}+\sum_{i=1}^{n}i^{2} & = & \sum_{k=1}^{n}k\left(k\right)+\sum_{k=1}^{n}\left(k\right)\left(2n-2k+1\right)+\sum_{k=1}^{n}k\left(k\right)\\
3\sum_{i=1}^{n}i^{2} & = & \sum_{k=1}^{n}k\left(2n-2k+1+2k\right)\\
 & = & \sum_{k=1}^{n}k\left(2n+1\right)\\
 & = & \left(2n+1\right)\sum_{k=1}^{n}k
\end{eqnarray*}
But we know that $\sum_{k=1}^{n}k=\frac{1}{2}\left(n\right)\left(n+1\right)$.
So we have
\begin{eqnarray*}
3\sum_{i=1}^{n}i^{2} & = & \frac{1}{2}\left(n\right)\left(n+1\right)\left(2n+1\right)\\
\sum_{i=1}^{n}i^{2} & = & \boxed{\dfrac{n\left(n+1\right)\left(2n+1\right)}{6}}
\end{eqnarray*}
as desired.

Notice how quickly we arrived at precisely the same conclusion using relatively primitive block drawing, without even having to recourse to sum notation, which was surprisingly clunky this time around.