Evaluate:
- \(\sum_{r=1}^{n}r(r+1)(r+3)\);
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 \[\begin{align*} \sum_{r=1}^{n}r(r+1)(r+3) &= \frac{n(n+1)[3n(n+1)+8(2n+1)+18]}{12} \\ &= \frac{n(n+1)(3n^2+19n+26)}{12} \\ &= \frac{n(n+1)(n+2)(3n+13)}{12}. \end{align*}\]- \(\sum_{r=1}^n \dfrac{1}{r(r+1)(r+3)}\).
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 \(= \dfrac{7}{36}-\dfrac{13}{360} = \dfrac{19}{120}\).
