The greatest integer lessthan or equal to the sum of first 100 terms of the sequence $\frac{1}{3},\frac{5}{9},\frac{19}{27},\frac{65}{81}\dots \dots \dots \dots =$

$$\begin{gathered} \mathrm{S}=\frac{1}{3}+\frac{5}{9}+\frac{19}{27}+\frac{65}{81}\dots \dots \dots .100 \mathrm{terms} \\ =\left(1-\frac{2}{3}\right)+\left(1-\frac{4}{9}\right)+\left(1-\frac{8}{27}\right)+\left(1-\frac{16}{81}\right)+\dots \dots .100 \mathrm{terms} \end{gathered}$$             $$\begin{gathered} \begin{array}{l}=100-\left[\left(\frac{2}{3}\right)+{\left(\frac{2}{3}\right)}^2+{\left(\frac{2}{3}\right)}^3+\dots \dots .\right] \\ =100-\frac{2}{3}\left[1+\frac{2}{3}+{\left(\frac{2}{3}\right)}^2+\dots \dots .\right]100\text{ terms }\\ =100-\frac{2}{3}\left[\frac{1-{\left(\frac{2}{3}\right)}^{100}}{1-\left(\frac{2}{3}\right)}\cdot \right] \\ =100-\frac{2}{3}\left[1-{\left(\frac{2}{3}\right)}^{100}\right]\times 3 \\ =100-2+2{\left(\frac{2}{3}\right)}^{100} \\ =98+2{\left(\frac{2}{3}\right)}^{100}\left[\mathrm{S}\right]=98\end{array} \end{gathered}$$