Find an expression for the sum of $r^2$

Find an expression for the sum of $r^2$

Question

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 $\sum_{r=0}^n r^2=\frac{1}{6}n(n+1)(2n+1).$ 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 $\sum_{r=0}^n r^3=\frac{1}{4}n^2(n+1)^2.$

Solution

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

$$\begin{align} n=-1 \qquad 1&=-p+q-r+s, \\ n=0 \qquad 1&=s, \\ n=1 \qquad 2&=p+q+r+s, \\ n=2 \qquad 6&= 8p+4q+2r+s. \end{align}$$

Equation

$(2)$

tells us that

$s=1$

, and so the other three equations become

$$\begin{align} -p+q-r&=0, \\ p+q+r&=1, \\ 8p+4q+2r&=5. \end{align}$$

Adding equations

$(1pr)$

and

$(2pr)$

, we find that

$$2q=1,$$

and so we have that

$$q=\frac{1}{2}.$$

Substituting this into our equations leaves us with two remaining;

$$\begin{align} p+r&=\frac{1}{2}, \\ 8p+2r&=3. \end{align}$$

Equation $(1prpr)$ tells us that $r=\dfrac{1}{2}-p$, and substituting this into equation $(2prpr)$ 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 $3$?

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

$$\begin{align} n=-2 \qquad -8&=16a-8b+4c-2d+e, \\ n=-1 \qquad -9&=a-b+c-d+e, \\ n=0 \qquad -9&=e, \\ n=1 \qquad -8 &=a+b+c+d+e, \\ n=2 \qquad \quad 0&=16a+8b+4c+2d+e. \end{align}$$

Equation

$(a4)$

tells us that

$e=-9$

, and so we can substitute this value into the remaining four equations, which gives us

$$\begin{align} 16a-8b+4c-2d&=1, \\ a-b+c-d&=0, \\ a+b+c+d&=1, \\ 16a+8b+4c+2d&=9. \end{align}$$

Examining equations

$(b1)$

and

$(b4)$

we see that they look similar, and so we can add them to find that

$$\begin{equation} 32a+8c=10 . \end{equation}$$

Adding equation

$(b2)$

and

$(b3)$

we have that

$$\begin{equation} 2a+2c=1 . \end{equation}$$

Rearranging this gives

$2c=1-2a$

, which we substitute into equation

$(c1)$

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

$(b1)$

,

$(b2)$

,

$(b3)$

and

$(b4)$

gives us

$$\begin{align} b+d=\frac{1}{2}, \\ 8b+2d=4. \end{align}$$

Equation $(d1)$ rearranges to $d=\frac{1}{2}-b$, which when substituted into equation $(d2)$ 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.