To solve the given series, we need to find the sum of the following expression:

\[ G = \frac{2}{1 \cdot 2 \cdot 3} + \frac{2}{2 \cdot 3 \cdot 4} + \frac{2}{3 \cdot 4 \cdot 5} + \cdots + \frac{2}{18 \cdot 19 \cdot 20} \]

Let's denote the general term of the series as:

\[ a_n = \frac{2}{n(n+1)(n+2)} \]

We can simplify \( a_n \) using partial fraction decomposition. We want to express \( a_n \) as:

\[ \frac{2}{n(n+1)(n+2)} = \frac{A}{n} + \frac{B}{n+1} + \frac{C}{n+2} \]

Multiplying both sides by \( n(n+1)(n+2) \) gives:

\[ 2 = A(n+1)(n+2) + Bn(n+2) + Cn(n+1) \]

Expanding and combining like terms, we get:

\[ 2 = A(n^2 + 3n + 2) + B(n^2 + 2n) + C(n^2 + n) \]
\[ 2 = (A + B + C)n^2 + (3A + 2B + C)n + 2A \]

For the equation to hold for all \( n \), the coefficients of \( n^2 \), \( n \), and the constant term must match on both sides. Therefore, we have the system of equations:

1. \( A + B + C = 0 \)
2. \( 3A + 2B + C = 0 \)
3. \( 2A = 2 \)

From the third equation, we find:

\[ A = 1 \]

Substituting \( A = 1 \) into the first and second equations:

\[ 1 + B + C = 0 \implies B + C = -1 \]
\[ 3(1) + 2B + C = 0 \implies 3 + 2B + C = 0 \implies 2B + C = -3 \]

Solving these two equations:

\[ B + C = -1 \]
\[ 2B + C = -3 \]

Subtract the first equation from the second:

\[ (2B + C) - (B + C) = -3 - (-1) \]
\[ B = -2 \]

Substitute \( B = -2 \) back into \( B + C = -1 \):

\[ -2 + C = -1 \]
\[ C = 1 \]

So, the partial fractions are:

\[ \frac{2}{n(n+1)(n+2)} = \frac{1}{n} - \frac{2}{n+1} + \frac{1}{n+2} \]

Thus, the series becomes:

\[ G = \left( \frac{1}{1} - \frac{2}{2} + \frac{1}{3} \right) + \left( \frac{1}{2} - \frac{2}{3} + \frac{1}{4} \right) + \left( \frac{1}{3} - \frac{2}{4} + \frac{1}{5} \right) + \cdots + \left( \frac{1}{18} - \frac{2}{19} + \frac{1}{20} \right) \]

Notice that most terms cancel out, leaving:

\[ G = 1 - \frac{1}{19} - \frac{1}{20} \]

Therefore, the sum \( G \) is:

\[ G = 1 - \frac{1}{19} - \frac{1}{20} = 1 - \frac{20 + 19}{19 \cdot 20} = 1 - \frac{39}{380} = 1 - \frac{39}{380} = \frac{380 - 39}{380} = \frac{341}{380} \]

So, the sum of the series is:

\[ G = \frac{341}{380} \]