Can we sum r(r+1)(r+3) for r from 1 to n?

Can we sum $r(r+1)(r+3)$ for $r$ from $1$ to $n$ ?

Question

Evaluate:

$\sum_{r=1}^{n}r(r+1)(r+3)$ ;

$\sum_{r=1}^n \dfrac{1}{r(r+1)(r+3)}$ .

Suggestion

Evaluate:

$\sum_{r=1}^n \dfrac{1}{r(r+1)(r+3)}$ .

How can we rewrite $\dfrac{1}{r(r+1)(r+3)}$ in a way that might be useful? Could we use partial fractions?

Or alternatively, if we write the numerator as $r+1-r$ , can we now split the fraction into two fractions which both have terms that cancel?

Solution

We multiply the brackets out, giving

\begin{align*} \sum_{r=1}^{n}r(r+1)(r+3) &= \sum_{r=1}^{n}(r^3+4r^2+3r) \\ &= \sum_{r=1}^{n}r^3+4\sum_{r=1}^{n}r^2+3\sum_{r=1}^{n}r. \end{align*}

Since we know $\sum_{r=1}^{n}r^3=\frac{n^2(n+1)^2}{4}, \sum_{r=1}^{n}r^2=\frac{n(n+1)(2n+1)}{6}, \sum_{r=1}^{n}r=\frac{n(n+1)}{2},$ we see that $\sum_{r=1}^{n}r(r+1)(r+3)=\frac{n^2(n+1)^2}{4}+4\frac{n(n+1)(2n+1)}{6}+3\frac{n(n+1)}{2},$ so $\sum_{r=1}^{n}r(r+1)(r+3)=\frac{3n^2(n+1)^2+8n(n+1)(2n+1)+18n(n+1)}{12}.$

We can now take out the common factor

$n(n+1)$

, to get

\begin{align*} \sum_{r=1}^{n}r(r+1)(r+3) &= \frac{n(n+1)[3n(n+1)+8(2n+1)+18]}{12} \\ &= \frac{n(n+1)(3n^2+19n+26)}{12} \\ &= \frac{n(n+1)(n+2)(3n+13)}{12}. \end{align*}

This looks like a

telescoping sum

; our only problem is that we have three numbers in the denominator. We can deal with that by rewriting the

$1$

in the numerator as

$r+1-r$

, so

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

Now we will compute the two sums separately. For the first one, we have

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

Since

\sum_{r=1}^{n}\frac{1}{r+3}=\sum_{r=4}^{n+3}\frac{1}{r},

we have

\begin{align*} \sum_{r=1}^{n} \frac{1}{r(r+3)} &= \frac{1}{3}\left(\sum_{r=1}^{n}\frac{1}{r}-\sum_{r=4}^{n+3}\frac{1}{r}\right) \\ &= \frac{1}{3}\left(1+\frac{1}{2}+\frac{1}{3}+\sum_{r=4}^{n}\frac{1}{r}-\sum_{r=4}^{n}\frac{1}{r}-\frac{1}{n+1}-\frac{1}{n+2}-\frac{1}{n+3}\right) \\ &= \frac{1}{3}\left(1+\frac{1}{2}+\frac{1}{3}-\frac{1}{n+1}-\frac{1}{n+2}-\frac{1}{n+3}\right). \end{align*}

For the second one we will try a similar approach. We have $\sum_{r=1}^n \frac{1}{(r+1)(r+3)} = \frac{1}{2}\sum_{r=1}^{n}\frac{(r+3)-(r+1)}{(r+1)(r+3)}=\frac{1}{2}\sum_{r=1}^{n}\left(\frac{1}{r+1}-\frac{1}{r+3}\right).$ Similarly to before, we have $\sum_{r=1}^n \frac{1}{(r+1)(r+3)}=\frac{1}{2}\left(\frac{1}{2}+\frac{1}{3}-\frac{1}{n+2}-\frac{1}{n+3}\right).$

By combining the results for the two sums, we get

\begin{align*} &\sum_{r=1}^n \frac{1}{r(r+1)(r+3)} \\ & \qquad = \frac{1}{3}\left(1+\frac{1}{2}+\frac{1}{3}-\frac{1}{n+1}-\frac{1}{n+2}-\frac{1}{n+3}\right)-\frac{1}{2}\left(\frac{1}{2}+\frac{1}{3}-\frac{1}{n+2}-\frac{1}{n+3}\right) \\ & \qquad = \frac{1}{3}+\frac{1}{6}+\frac{1}{9}-\frac{1}{4}-\frac{1}{6}-\frac{1}{3(n+1)}+\left(\frac{1}{n+2}+\frac{1}{n+3}\right)\left(\frac{1}{2}-\frac{1}{3}\right) \\ & \qquad = \frac{7}{36}-\frac{1}{3(n+1)}+\frac{2n+5}{6(n+2)(n+3)} \\ & \qquad = \frac{7}{36}+\frac{(2n+5)(n+1)-2(n+2)(n+3)}{6(n+1)(n+2)(n+3)} \\ & \qquad = \frac{7}{36}-\frac{3n+7}{6(n+1)(n+2)(n+3)}. \end{align*}

We could alternatively write $\dfrac{1}{r(r+1)(r+3)} = \dfrac{A}{r}+\dfrac{B}{r+1}+\dfrac{C}{r+3}$ . Finding the constants in the usual way gives $\dfrac{1}{r(r+1)(r+3)} = \dfrac{1/3}{r}-\dfrac{1/2}{r+1}+\dfrac{1/6}{r+3}.$ Note that $\dfrac{1}{3}-\dfrac{1}{2}+\dfrac{1}{6} = 0$ , which means that most of the terms cancel (as before). This leaves us with $\sum_{r=1}^n \frac{1}{r(r+1)(r+3)} = \dfrac{11}{18}-\dfrac{5}{12}-\dfrac{1}{2(n+1)}+\dfrac{1}{6}\left(\dfrac{1}{n+1}+\dfrac{1}{n+2} + \dfrac{1}{n+3}\right)$ $= \frac{7}{36}-\frac{3n+7}{6(n+1)(n+2)(n+3)}.$

Check: when $n = 2$ , LHS $= \dfrac{1}{8}+ \dfrac{1}{30}=\dfrac{19}{120}$ , and RHS $= \dfrac{7}{36}-\dfrac{13}{360} = \dfrac{19}{120}$ .