The first triangle number is $1$; the second is $1 + 2 = 3$; the third is $1 + 2 + 3 = 6$; and so on. Find the sum of the first $100$ triangle numbers.
1. The $n$-th triangular number is given by the formula:
   \[
   T_n = \frac{n(n+1)}{2}
   \]
   This formula can be derived from the sum of the first $n$ natural numbers.

2. We need to find the sum of the first 100 triangular numbers:
   \[
   \sum_{n=1}^{100} T_n = \sum_{n=1}^{100} \frac{n(n+1)}{2}
   \]

3. We can simplify the sum by factoring out the constant $\frac{1}{2}$:
   \[
   \sum_{n=1}^{100} \frac{n(n+1)}{2} = \frac{1}{2} \sum_{n=1}^{100} n(n+1)
   \]

4. Next, we expand the product $n(n+1)$:
   \[
   n(n+1) = n^2 + n
   \]

5. Substituting this back into the sum, we get:
   \[
   \frac{1}{2} \sum_{n=1}^{100} (n^2 + n)
   \]

6. We can separate the sum into two parts:
   \[
   \frac{1}{2} \left( \sum_{n=1}^{100} n^2 + \sum_{n=1}^{100} n \right)
   \]

7. We use the formulas for the sum of the first $n$ natural numbers and the sum of the squares of the first $n$ natural numbers:
   \[
   \sum_{n=1}^{100} n = \frac{100(100+1)}{2} = 5050
   \]
   \[
   \sum_{n=1}^{100} n^2 = \frac{100(100+1)(2 \cdot 100 + 1)}{6} = \frac{100 \cdot 101 \cdot 201}{6} = 338350
   \]

8. Substituting these values back into our sum, we get:
   \[
   \frac{1}{2} \left( 338350 + 5050 \right) = \frac{1}{2} \cdot 343400 = 171700
   \]

Therefore, the sum of the first 100 triangular numbers is:
\[
\boxed{171700}
\]