Question

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

Answer 1

Answer: 98

$\frac{1}{3}+\frac{5}{9}+\frac{19}{27}+\frac{65}{81}+\dots$
$\left(1-\frac{2}{3}\right)+\left(1-\frac{4}{9}\right)+\left(1-\frac{8}{27}\right)+\left(1-\frac{16}{81}\right)\dots .100$ terms
$100-\left[\frac{2}{3}+{\left(\frac{2}{3}\right)}^2+\dots \right]$
$100-\frac{\frac{2}{3}\left(1-{\left(\frac{2}{3}\right)}^{100}\right)}{1-\frac{2}{3}}$
$100-2\left(1-{\left(\frac{2}{3}\right)}^{100}\right)$
$S=98+2{\left(\frac{2}{3}\right)}^{100}$
$\Rightarrow [S]=98$