2025 ASRJC P1 Q6

It is given that \(\sum_{r=1}^{n} \frac{7r+4}{r(r+1)(r+2)} = \frac{9}{2} - \frac{2}{n+1} - \frac{5}{n+2}\).

1. Find \(\sum_{r=7}^{2n} \frac{7r-3}{r^3-r}\) giving your answers in terms of \(n\).[4]
2. Show algebraically that \((r+1)^3 > r(r+1)(r+2)\) for all positive integers \(r\).[2]
3. Hence show that \(\sum_{r=1}^{n} \frac{7r+4}{(r+1)^3} < \frac{9}{2}\).[3]