Answer: (a) \(\dfrac14 n(n+1)(n-1)(n+2)\)

(b) \(\dfrac32-\dfrac2n+\dfrac1{n+1}\)

(c) \(\dfrac32\)

(a)

\(\sum_{r=1}^{n}(r^3-r)=\sum_{r=1}^{n}r^3-\sum_{r=1}^{n}r.\)

Using MF19,

\(\sum r^3=\left(\frac{n(n+1)}2\right)^2,\qquad \sum r=\frac{n(n+1)}2.\)

Hence

\(\left(\frac{n(n+1)}2\right)^2-\frac{n(n+1)}2=\frac14n(n+1)\{n(n+1)-2\}.\)

Since \(n(n+1)-2=n^2+n-2=(n-1)(n+2)\), the answer is

\(\frac14 n(n+1)(n-1)(n+2).\)

(b)

Since \(r^3-r=r(r-1)(r+1)\), write

\(\frac{r+3}{r(r-1)(r+1)}=\frac{A}{r-1}+\frac{B}{r}+\frac{C}{r+1}.\)

Multiplying by \(r(r-1)(r+1)\) gives

\(r+3=Ar(r+1)+B(r-1)(r+1)+Cr(r-1).\)

Solving gives \(A=2, B=-3, C=1\), so

\(\frac{r+3}{r^3-r}=\frac2{r-1}-\frac3r+\frac1{r+1}.\)

Therefore

\(\sum_{r=2}^{n}\left(\frac2{r-1}-\frac3r+\frac1{r+1}\right)\)

has cancelling terms, leaving

\(\frac32-\frac2n+\frac1{n+1}.\)

(c)

From part (b),

\(\sum_{r=2}^{n}\frac{r+3}{r^3-r}=\frac32-\frac2n+\frac1{n+1}.\)

Letting \(n\to\infty\) gives

\(\frac32.\)

Mark scheme