To find the sum \( F \) of the series \( 6 + 16 + 30 + 48 + \ldots + 19600 + 19998 \), we first need to identify the pattern or formula for the \( n \)-th term of the series.

Let's examine the given terms:
- The 1st term is \( 6 \).
- The 2nd term is \( 16 \).
- The 3rd term is \( 30 \).
- The 4th term is \( 48 \).

We can try to find a pattern by looking at the differences between consecutive terms:
- \( 16 - 6 = 10 \)
- \( 30 - 16 = 14 \)
- \( 48 - 30 = 18 \)

The differences themselves are increasing by 4 each time:
- \( 14 - 10 = 4 \)
- \( 18 - 14 = 4 \)

This suggests that the series might be quadratic in nature. Let's assume the \( n \)-th term of the series can be expressed as \( a_n = an^2 + bn + c \).

Using the first few terms to set up equations:
1. For \( n = 1 \): \( a(1)^2 + b(1) + c = 6 \) \(\Rightarrow a + b + c = 6 \)
2. For \( n = 2 \): \( a(2)^2 + b(2) + c = 16 \) \(\Rightarrow 4a + 2b + c = 16 \)
3. For \( n = 3 \): \( a(3)^2 + b(3) + c = 30 \) \(\Rightarrow 9a + 3b + c = 30 \)

We now have a system of three equations:
1. \( a + b + c = 6 \)
2. \( 4a + 2b + c = 16 \)
3. \( 9a + 3b + c = 30 \)

Subtract the first equation from the second:
\[ (4a + 2b + c) - (a + b + c) = 16 - 6 \]
\[ 3a + b = 10 \quad \text{(Equation 4)} \]

Subtract the second equation from the third:
\[ (9a + 3b + c) - (4a + 2b + c) = 30 - 16 \]
\[ 5a + b = 14 \quad \text{(Equation 5)} \]

Subtract Equation 4 from Equation 5:
\[ (5a + b) - (3a + b) = 14 - 10 \]
\[ 2a = 4 \]
\[ a = 2 \]

Substitute \( a = 2 \) back into Equation 4:
\[ 3(2) + b = 10 \]
\[ 6 + b = 10 \]
\[ b = 4 \]

Substitute \( a = 2 \) and \( b = 4 \) back into the first equation:
\[ 2 + 4 + c = 6 \]
\[ 6 + c = 6 \]
\[ c = 0 \]

So, the \( n \)-th term of the series is:
\[ a_n = 2n^2 + 4n \]

To find the sum of the series up to the term where \( a_n = 19998 \), we need to solve for \( n \):
\[ 2n^2 + 4n = 19998 \]
\[ n^2 + 2n - 9999 = 0 \]

Using the quadratic formula \( n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ n = \frac{-2 \pm \sqrt{4 + 4 \cdot 9999}}{2} \]
\[ n = \frac{-2 \pm \sqrt{4 + 39996}}{2} \]
\[ n = \frac{-2 \pm \sqrt{40000}}{2} \]
\[ n = \frac{-2 \pm 200}{2} \]
\[ n = \frac{198}{2} \quad \text{or} \quad n = \frac{-202}{2} \]
\[ n = 99 \quad \text{or} \quad n = -101 \]

Since \( n \) must be positive, \( n = 99 \).

Now, we need to find the sum of the series from \( n = 1 \) to \( n = 99 \):
\[ F = \sum_{n=1}^{99} (2n^2 + 4n) \]

This can be split into two separate sums:
\[ F = 2 \sum_{n=1}^{99} n^2 + 4 \sum_{n=1}^{99} n \]

Using the formulas for the sum of squares and the sum of the first \( n \) natural numbers:
\[ \sum_{n=1}^{k} n^2 = \frac{k(k+1)(2k+1)}{6} \]
\[ \sum_{n=1}^{k} n = \frac{k(k+1)}{2} \]

For \( k = 99 \):
\[ \sum_{n=1}^{99} n^2 = \frac{99 \cdot 100 \cdot 199}{6} = 328350 \]
\[ \sum_{n=1}^{99} n = \frac{99 \cdot 100}{2} = 4950 \]

Therefore:
\[ F = 2 \cdot 328350 + 4 \cdot 4950 \]
\[ F = 656700 + 19800 \]
\[ F = 676500 \]

So, the sum \( F \) of the series is \( 676500 \).