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?
 A: Using Summation by Parts, with $f_r=r^2$ and $g_r=r$, we have
$$\begin{align}
\sum_{r=1}^n r^2&=(n+1)^3-1-\sum_{r=1}^n (r+1)\left((r+1)^2-r^2\right)\\\\
&=(n+1)^3-1-\sum_{r=1}^n (r+1)\left(2r+1\right)\\\\
&=(n+1)^3-1-2\sum_{r=1}^n r^2-3\sum_{r=1}^{n}r-\sum_{r=1}^{n}1\\\\
3\sum_{r=1}^n r^2&=(n+1)^3-1-3\sum_{r=1}^{n}r-\sum_{r=1}^{n}1\\\\
\sum_{r=1}^n r^2&=\frac{n(n+1)}{2}\frac{2(n+2)}{3}-\sum_{r=1}^{n}r\\\\
&=\frac{n(n+1)}{2}\left(\frac{2(n+2)}{3}-1\right)\\\\
&=\frac{n(n+1)}{2}\frac{2n+1}{3}
\end{align}$$

ALTERNATIVE SUMMATION BY PARTS
We can instead use the Newton Series for summation by parts.  Here, we let $f_k=g_k=k$ and write
$$\begin{align}
\sum_{k=1}^{n}k^2&=n\left(\frac{n(n+1)}{2}\right)-\sum_{k=0}^{n-1}\sum_{\ell=0}^{k} k\\\\
&=n\left(\frac{n(n+1)}{2}\right)-\sum_{k =0}^{n-1}\frac{k(k+1)}{2}\\\\
&=(n+1)\left(\frac{n(n+1)}{2}\right)-\frac12\sum_{k =1}^n k(k+1)\\\\
\frac32\sum_{k=1}^{n}k^2&=(n+1)\left(\frac{n(n+1)}{2}\right)-\frac12\sum_{k =1}^n k\\\\
\sum_{k=1}^{n}k^2&=\frac23 (n+1)\left(\frac{n(n+1)}{2}\right)-\frac13\sum_{k =1}^n\,k\\\\
&=\frac13(2n+1)\left(\frac{n(n+1)}{2}\right)
\end{align}$$
again as expected!
A: 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.

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