The greatest integer less than or equal to the sum of first 100...

Let $S = \dfrac{1}{3} + \dfrac{5}{9} + \dfrac{19}{27} + ⋯$ upto $100$ terms

$S = \dfrac{3 - 2}{{3}^{1}} + \dfrac{{3}^{2} - {2}^{2}}{{3}^{2}} + \dfrac{{3}^{3} - {2}^{3}}{{3}^{3}} + . . . .$ upto $100$ terms

So, ${T}_{n} = 1 - {\left(\dfrac{2}{3}\right)}^{n}$$⇒ {S}_{100} = 100 - \left[\dfrac{\dfrac{2}{3}\left[1 - {\left(\dfrac{2}{3}\right)}^{100}\right]}{1 - \dfrac{2}{3}}\right]$

$= 98 + 2{\left(\dfrac{2}{3}\right)}^{100}$ $⇒ \left[{S}_{100}\right] = 98$