(a) Show that, for r>0r > 0r>0 r+2r(r+1)−r+3(r+1)(r+2)=r+4r(r+1)(r+2)\begin{align*} \frac{r + 2}{r(r + 1)} - \frac{r + 3}{(r + 1)(r + 2)} = \frac{r + 4}{r(r + 1)(r + 2)} \end{align*}r(r+1)r+2−(r+1)(r+2)r+3=r(r+1)(r+2)r+4 (2) (b) Hence show that ∑r=1nr+4r(r+1)(r+2)=c(n+a)(n+b)n(n+1)(n+2)\begin{align*} \sum_{r=1}^{n} \frac{r + 4}{r(r + 1)(r + 2)} = \frac{c(n + a)(n + b)}{n(n + 1)(n + 2)} \end{align*}r=1∑nr(r+1)(r+2)r+4=n(n+1)(n+2)c(n+a)(n+b) where aaa, bbb and ccc are integers to be determined. (4)
