Review question

# Find an expression for the sum of $r^2$ Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R6143

## Solution

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 $$$32a+8c=10 \label{c1}.$$$ Adding equation $\eqref{b2}$ and $\eqref{b3}$ we have that $$$2a+2c=1 \label{c2}.$$$ 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.