This sum telescopes. We are left with only the \(-a_0\) from the first equation and the \(a_n\) from the last equation. Putting this all together we have \(-a_0 + a_n = \frac{n(n+1)}{2}\) or \(a_n = \frac{n(n+1)}{2} + a_0\text{.}\) But we know that \(a_0 = 4\text{.}\) So the solution to the recurrence relation, subject to the initial condition is