summation - Sum of Squares in terms of Sum of Integers

We know that the sum of squares can be expressed as a multiple of the sum of integers as follows: $$\begin{align}
\sum_{r=1}^n r^2
&=\frac 16 n(n+1)(2n+1)\\
&=\frac {2n+1}3\cdot \frac {n(n+1)}2\\
&=\frac {2n+1}3\sum_{r=1}^nr\end{align}$$

Is there a simple direct proof to express the sum of squares as $\dfrac {2n+1}3$ multiplied by the sum of integers, without first deriving the formula for the sum of squares and then breaking it down as shown above?

Answer

New Solution

Have just found a proof which does not require the prior knowledge of the result of the sum of integers.

$$\begin{align}
\sum_{i=1}^n i^2&=\sum_{i=1}^n\sum_{j=1}^i(2j-1)&& ...(1)\\
\sum_{i=1}^n i^2&=\sum_{i=1}^n\sum_{j=1}^i 2(n-i)+1&& ...(2)\\
\sum_{i=1}^n i^2&=\sum_{i=1}^n\sum_{j=1}^i 2(i-j)+1&& ...(3)\\
(1)+(2)+(3):\\

3\sum_{i=1}^n i^2
&=\sum_{i=1}^n\sum_{j=1}^i (2j-1)+2(n-1)+1+2(i-j)+1\\
&=\sum_{i=1}^n\sum_{j=1}^i (2n+1)\\
&=(2n+1)\sum_{i=1}^n i\\
\sum_{i=1}^n i^2&=\frac{2n+1}3\sum_{i=1}^n i\qquad\blacksquare
\end{align}$$

This proof is a transcription of the diagrammatic proof of the same as shown on the wikipedia page here.

See also the nice diagrams in this solution here.

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Earlier post shown below

This is too long for a comment so it's being posted in the solution section.

A synthetic and rather cumbersome approach might be as follows:

$$\begin{align}

(2m+1)\sum_{r=1}^mr-(2m-1)\sum_{r=1}^{m-1}r
&=(2m+1)\frac {m(m+1)}2-(2m-1)\frac{m(m-1)}2\\
&=\frac m2\left[(2m+1)(m+1)-(2m-1)(m-1)\right]\\
&=3m^2\end{align}$$
Summing $m$ from $1$ to $n$ and telescoping LHS gives

$$(2n+1)\sum_{r=1}^nr=3\sum_{m=1}^nm^2\color{lightgray}{=3\sum_{r=1}^n r^2}\\
\sum_{r=1}^nr^2=\frac {2n+1}3\sum_{r=1}^nr\qquad\blacksquare$$