Question

The greatest integer less than or equal to the sum of first $100$ terms of the sequence $\frac 13, \frac 59, \frac {19}{27}, \frac {65}{81},…...$ is equal to ______.

Answer 1

$\frac 13, \frac 59, \frac {19}{27}, \frac {65}{81},…...$

$S= \displaystyle\sum_{r=1}^{100} (\frac {3^r−2^r}{3^r})$

$S=100−\frac 23 \frac {(1−(\frac 23)^{100})}{\frac 13}$

$S=98+2(\frac 23)^{100}$
$S = 98$

So, the answer is $98$.