The greatest integer less than or equal to the sum of first $$100$$ terms of the sequence $$\frac{1}{3}, \frac{5}{9}, \frac{19}{27}, \frac{65}{81}, \ldots$$ is equal to ______.

The sequence is $$\frac{1}{3}, \frac{5}{9}, \frac{19}{27}, \frac{65}{81}, \ldots$$ and the denominators follow $$3^1, 3^2, 3^3, 3^4, \ldots$$ while the numerators $$1, 5, 19, 65, \ldots$$ satisfy $$3^n - 2^n$$ since $$3^1 - 2^1 = 1$$, $$3^2 - 2^2 = 5$$, $$3^3 - 2^3 = 19$$, and $$3^4 - 2^4 = 65$$. Consequently, the general term can be written as $$a_n = \frac{3^n - 2^n}{3^n} = 1 - \left(\frac{2}{3}\right)^n$$.

To find the sum of the first 100 terms, we write $$S_{100} = \sum_{n=1}^{100}\Bigl[1 - \Bigl(\tfrac{2}{3}\Bigr)^n\Bigr] = 100 - \sum_{n=1}^{100}\Bigl(\tfrac{2}{3}\Bigr)^n$$. The geometric series sum is $$\sum_{n=1}^{100}\Bigl(\tfrac{2}{3}\Bigr)^n = \frac{\tfrac{2}{3}\bigl(1 - (\tfrac{2}{3})^{100}\bigr)}{1 - \tfrac{2}{3}} = 2\bigl(1 - (\tfrac{2}{3})^{100}\bigr)$$, which gives $$S_{100} = 100 - 2 + 2\Bigl(\tfrac{2}{3}\Bigr)^{100} = 98 + 2\Bigl(\tfrac{2}{3}\Bigr)^{100}$$.

Since $$\Bigl(\frac{2}{3}\Bigr)^{100}$$ is extremely small, essentially zero, we have $$S_{100} = 98 + 2\Bigl(\frac{2}{3}\Bigr)^{100}$$ which lies between 98 and 99. Therefore $$\lfloor S_{100}\rfloor = 98$$, and the answer is $$98$$.