1. **Problem A: Find a closed-form for $a_n$ and determine $a_{100}$.** Given $a_1=1$ and recursive relation $$a_{n+1} = a_n + n + 1,$$ we want to express $a_n$ without recursion. 2. Write out first few terms: $$a_2 = a_1 + 2 = 1 + 2 = 3,$$ $$a_3 = a_2 + 3 = 3 + 3 = 6,$$ $$a_4 = a_3 + 4 = 6 + 4 = 10.$$ 3. Notice pattern: $a_n$ looks like sum of first $n$ integers plus 1. 4. Unroll recursion: $$a_n = a_1 + (2 + 3 + 4 + \cdots + n),$$ since from $a_1$ to $a_n$ adds terms from 2 to $n$. 5. Sum inside is arithmetic sum: $$\sum_{k=2}^n k = \left(\sum_{k=1}^n k\right) - 1 = \frac{n(n+1)}{2} - 1.$$ 6. Substitute back: $$a_n = 1 + \left(\frac{n(n+1)}{2} - 1\right) = \frac{n(n+1)}{2}.$$ 7. So closed form is $$a_n = \frac{n(n+1)}{2}.$$ 8. Calculate $a_{100}$: $$a_{100} = \frac{100 \times 101}{2} = 50 \times 101 = 5050.$$