Step-by-Step Solution
Key Concept: Transform the recurrence relation into a telescoping form to evaluate sums involving operator differences.
Expand $((r+1)^2 - r^2)\phi(r) = (2r+1)\phi(r)$ and rearrange to obtain a telescoping sum: $\sum_{r=1}^{n}((r+1)^2\phi(r+1) - r^2\phi(r)) = -\sum_{r=1}^{n}(r+1)(n(r+1)^2\delta(n+1)-1^2\delta(1)) = (n+1)^2\phi(n+1) - \phi(1)$.
Correct Answer: 4