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.
