1. **Problem:** Calculate the sum $$\frac{1}{2 \times 6} + \frac{1}{4 \times 9} + \frac{1}{6 \times 12} + \ldots + \frac{1}{36 \times 57} + \frac{1}{38 \times 60}$$ 2. **Step 1:** Notice the pattern of terms: the numerator is always 1, and the denominators are products of two numbers where the first number increases by 2 starting from 2 and the second number increases by 3 starting from 6. General term: $$\frac{1}{(2n)(3n+3)}$$ for integers $n$ starting at 1. 3. **Step 2:** Simplify the denominator: $$ (2n)(3n+3) = 6n(n+1) $$ 4. **Step 3:** Use partial fraction decomposition: $$ \frac{1}{6n(n+1)} = \frac{A}{n} + \frac{B}{n+1} $$ Multiply both sides by $6n(n+1)$: $$ 1 = 6A(n+1) + 6Bn $$ Let $n = -1 \to$ left side 1, right side $6A(0) + 6B(-1) = -6B$\ So, $1 = -6B \implies B = -\frac{1}{6}$ Let $n = 0 \to 1 = 6A(1) + 0 \implies 1 = 6A \implies A = \frac{1}{6}$ 5. **Step 4:** Thus, $$ \frac{1}{6n(n+1)} = \frac{1}{6n} - \frac{1}{6(n+1)} $$ 6. **Step 5:** Sum from $n=1$ to $n=19$ (since $2 \times 38=76$, and the last term is $38 \times 60$, so $n=19$ for the last term): $$ \sum_{n=1}^{19} \left( \frac{1}{6n} - \frac{1}{6(n+1)} \right) = \frac{1}{6} \left( \sum_{n=1}^{19} \frac{1}{n} - \sum_{n=2}^{20} \frac{1}{n} \right) = \frac{1}{6} \left(1 - \frac{1}{20} \right) = \frac{1}{6} \times \frac{19}{20} = \frac{19}{120} $$ --- **Final answer for problem 1:** $$\boxed{\frac{19}{120}}$$