It is given that \(\sum\limits_{r=-1}^n r^2\) can be written in the form \(pn^3+qn^2+rn+s\), where \(p,q,r\) and \(s\) are numbers. By setting \(n=-1,0,1\) and \(2\), obtain four equations that must be satisfied by \(p,q,r\) and \(s\) and hence show that \[ \sum_{r=0}^n r^2=\frac{1}{6}n(n+1)(2n+1). \]

There are ways that we could work out that \(\sum\limits_{r=0}^n r^2\) will be a cubic function, even if we didn’t know the result stated above. You might like to look at Sum estimating to explore this idea.

Let’s substitute \(n=-1,0,1\) and \(2\) into the equation \(\sum_{r=-1}^n r^2=pn^3+qn^2+rn+s.\) We find \[\begin{align}\label{1} n=-1 \qquad 1&=-p+q-r+s, \\ \label{2} n=0 \qquad 1&=s, \\ \label{3} n=1 \qquad 2&=p+q+r+s, \\ \label{4} n=2 \qquad 6&= 8p+4q+2r+s. \end{align}\] Equation \(\eqref{2}\) tells us that \(s=1\), and so the other three equations become \[\begin{align}\label{1pr} -p+q-r&=0, \\ \label{2pr} p+q+r&=1, \\ \label{3pr} 8p+4q+2r&=5. \end{align}\] Adding equations \(\eqref{1pr}\) and \(\eqref{2pr}\), we find that \[ 2q=1, \] and so we have that \[ q=\frac{1}{2}. \] Substituting this into our equations leaves us with two remaining; \[\begin{align} \label{1prpr} p+r&=\frac{1}{2}, \\ \label{2prpr} 8p+2r&=3. \end{align}\]

Equation \(\eqref{1prpr}\) tells us that \(r=\dfrac{1}{2}-p\), and substituting this into equation \(\eqref{2prpr}\) tells us that \[ 8p+2\left(\frac{1}{2}-p\right)=3, \] which we rearrange to find that \[ 6p=2 \implies p=\frac{1}{3}. \] Therefore \[ r=\frac{1}{2}-p=\frac{1}{2}-\frac{1}{3}=\frac{1}{6}. \] We’ve calculated that \[ \sum_{r=-1}^n r^2=\frac{1}{3}n^3+\frac{1}{2}n^2+\frac{1}{6}n+1. \] Finally, we write \[ \sum_{r=0}^n r^2=\sum_{r=-1}^n r^2-1=\frac{1}{3}n^3+\frac{1}{2}n^2+\frac{1}{6}n=\frac{n}{6}(2n^2+3n+1)=\frac{n}{6}(2n+1)(n+1), \] as required.

Why do you think that the question gave us \(\sum\limits_{r=-1}^n r^2\) and asked us to substitute \(n=-1,0,1\) and \(2\), rather than \(\sum\limits_{r=0}^n r^2\) and \(n=0,1, 2\) and \(3\)?

Given that \(\sum\limits_{r=-2}^n r^3\) can be written in the form \(an^4+bn^3+cn^2+dn+e\), show similarly that \[ \sum_{r=0}^n r^3=\frac{1}{4}n^2(n+1)^2. \]

This time there are five unknowns, so we should substitute \(n=-2,-1,0,1\) and \(2\) into the equation \[ \sum_{r=-2}^n r^3=an^4+bn^3+cn^2+dn+e. \] Doing this yields the equations \[\begin{align}\label{a1} n=-2 \qquad -8&=16a-8b+4c-2d+e, \\ \label{a2} n=-1 \qquad -9&=a-b+c-d+e, \\ \label{a3} n=0 \qquad -9&=e, \\ \label{a4} n=1 \qquad -8 &=a+b+c+d+e, \\ \label{a5} n=2 \qquad \quad 0&=16a+8b+4c+2d+e. \end{align}\] Equation \(\eqref{a4}\) tells us that \(e=-9\), and so we can substitute this value into the remaining four equations, which gives us \[\begin{align} \label{b1} 16a-8b+4c-2d&=1, \\ \label{b2} a-b+c-d&=0, \\ \label{b3} a+b+c+d&=1, \\ \label{b4} 16a+8b+4c+2d&=9. \end{align}\] Examining equations \(\eqref{b1}\) and \(\eqref{b4}\) we see that they look similar, and so we can add them to find that \[\begin{equation} 32a+8c=10 \label{c1}. \end{equation}\] Adding equation \(\eqref{b2}\) and \(\eqref{b3}\) we have that \[\begin{equation} 2a+2c=1 \label{c2}. \end{equation}\] Rearranging this gives \(2c=1-2a\), which we substitute into equation \(\eqref{c1}\) to see that \[ 32a+4(1-2a)=10 \implies 24a=6, \] or \[ a=\frac{1}{4} \implies c=\frac{1}{4}. \] Substituting these values into the equations \(\eqref{b1}\), \(\eqref{b2}\), \(\eqref{b3}\) and \(\eqref{b4}\) gives us \[\begin{align}\label{d1} b+d=\frac{1}{2}, \\ \label{d2} 8b+2d=4. \end{align}\]

Equation \(\eqref{d1}\) rearranges to \(d=\frac{1}{2}-b\), which when substituted into equation \(\eqref{d2}\) tells us that \[ 8b+2\left(\frac{1}{2}-b\right)=4 \implies 6b=3 \implies b=\frac{1}{2}, d=0. \] So we’ve found that \[ \sum_{r=-2}^n r^3=\frac{1}{4}n^4+\frac{1}{2}n^3+\frac{1}{4}n^2-9. \] Therefore, \[ \sum_{r=0}^n r^3=\sum_{r=-2}^n r^3+9=\frac{1}{4}n^4+\frac{1}{2}n^3+\frac{1}{4}n^2=\frac{n^2}{4}(n^2+2n+1)=\frac{n^2}{4}(n+1)^2, \] as required.