The greatest integer less than or equal to th | vedclass

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Question

$\frac{1}{3}+\frac{5}{9}+\frac{19}{27}+\frac{65}{81}+\ldots$

$\left(1-\frac{2}{3}\right)+\left(1-\frac{4}{9}\right)+\left(1-\frac{8}{27}\right)+\le

Vedclass · Accepted Answer

$98$

Vedclass · Answer

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

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

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