Solution
= Solution
Unroll the <recurrence relation> down to its <initial condition>:
$$
T(n)=T(0)+\sum_{j=1}^{n}j=1+\frac{n(n+1)}2.
$$
Thus
$$
\boxed{T(n)=1+\frac{n(n+1)}2=O(n^2)}.
$$
Indeed $T(n)/n^2\to1/2$, so the sharper growth classification is $\Theta(n^2)$ in <big theta notation>.