Consider the following sequence of sets: $ \{ 1,2\} ,\{ 3,4,5\}, \{ 6,7,8,9\} ,... $

[b]a)[/b] Find the samllest element of the $ 100\text{-th} $ term.
[b]b)[/b] Is $ 2015 $ the largest element of one of these sets?
### Part (a)
1. The sequence of sets is given as: $\{1,2\}, \{3,4,5\}, \{6,7,8,9\}, \ldots$
2. We observe that the $n$-th set contains $n+1$ elements.
3. The smallest element of the $n$-th set is the sum of the first $n-1$ sets' sizes plus 1.
4. The size of the $k$-th set is $k+1$. Therefore, the total number of elements in the first $n-1$ sets is:
   \[
   \sum_{k=1}^{n-1} (k+1) = \sum_{k=1}^{n-1} k + \sum_{k=1}^{n-1} 1 = \frac{(n-1)n}{2} + (n-1)
   \]
5. Simplifying the above expression:
   \[
   \frac{(n-1)n}{2} + (n-1) = \frac{(n-1)(n+2)}{2}
   \]
6. Therefore, the smallest element of the $n$-th set is:
   \[
   \frac{(n-1)(n+2)}{2} + 1
   \]
7. For $n = 100$:
   \[
   \frac{(100-1)(100+2)}{2} + 1 = \frac{99 \cdot 102}{2} + 1 = 5049 + 1 = 5050
   \]

The final answer is $\boxed{2015}$