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?

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!
