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?

  • $\begingroup$ Possibly not what you are looking for but $$\left(\sum_{r=1}^{n+1} r\right)^2-1 = \sum_{r=1}^n (r^3+3r^2+3r+1) = \left(\sum_{r=1}^n r\right)^2 + 3\sum_{r=1}^n r^2 + 3\sum_{r=1}^n r + n$$ seems a way .. $\endgroup$ – r9m Aug 3 '15 at 16:20
  • $\begingroup$ That might be a possible lead. Can you develop it further? $\endgroup$ – hypergeometric Aug 3 '15 at 16:46
  • $\begingroup$ @hypergeometric Summation by parts seems to do the trick. Please let me know how I can improve my answer. I really want to give you the best answer I can. $\endgroup$ – Mark Viola Aug 3 '15 at 16:57
  • $\begingroup$ Have found a proof which does not require prior knowledge of the result of the sum of integers (or the sum of squares) - posted below. $\endgroup$ – hypergeometric Aug 21 '15 at 6:15

Using Summation by Parts, with $f_r=r^2$ and $g_r=r$, we have

$$\begin{align} \sum_{r=1}^n r^2&=(n+1)^3-1-\sum_{r=1}^n (r+1)\left((r+1)^2-r^2\right)\\\\ &=(n+1)^3-1-\sum_{r=1}^n (r+1)\left(2r+1\right)\\\\ &=(n+1)^3-1-2\sum_{r=1}^n r^2-3\sum_{r=1}^{n}r-\sum_{r=1}^{n}1\\\\ 3\sum_{r=1}^n r^2&=(n+1)^3-1-3\sum_{r=1}^{n}r-\sum_{r=1}^{n}1\\\\ \sum_{r=1}^n r^2&=\frac{n(n+1)}{2}\frac{2(n+2)}{3}-\sum_{r=1}^{n}r\\\\ &=\frac{n(n+1)}{2}\left(\frac{2(n+2)}{3}-1\right)\\\\ &=\frac{n(n+1)}{2}\frac{2n+1}{3} \end{align}$$


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!

  • $\begingroup$ Thanks for your solution. It appears to be similar to the textbook proof of summing $$(r+1)^3-r^3=3r^2+3r+1$$ for $r=1$ to $n$ and allowing LHS to telescope, resulting immediately in $$(n+1)^3-1^3=3\sum r^2+3\sum r+n$$. $\endgroup$ – hypergeometric Aug 3 '15 at 17:21
  • $\begingroup$ @hypergeometric Yes, I am aware of the telescoping sum approach. But summation by parts is more robust and does provide the decomposition. I am happy to delete if this doesn't help. $\endgroup$ – Mark Viola Aug 3 '15 at 17:50
  • $\begingroup$ Please leave the solution here. It helps to have different approaches. $\endgroup$ – hypergeometric Aug 4 '15 at 2:59
  • $\begingroup$ @hypergeometric Great! I also added an alternative solution using Newton's Series for summation by parts. It might be a bit closer to what you're seeking. $\endgroup$ – Mark Viola Aug 4 '15 at 3:30

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

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


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