The greatest integer less than or equal to th | vedclass

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(B) The $n$-th term of the sequence is $a_n = \frac{3^n - 2^n}{3^n} = 1 - \left(\frac{2}{3}\right)^n$.The sum of the first $100$ terms is $S_{100} = \

Vedclass · Accepted Answer

$98$

Vedclass · Answer

(B) The $n$-th term of the sequence is $a_n = \frac{3^n - 2^n}{3^n} = 1 - \left(\frac{2}{3}\right)^n$.The sum of the first $100$ terms is $S_{100} = \sum_{n=1}^{100} \left(1 - \left(\frac{2}{3}\right)^n\right)$.$S_{100} = \sum_{n=1}^{100} 1 - \sum_{n=1}^{100} \left(\frac{2}{3}\right)^n$.$S_{100} = 100 - \left[ \frac{2}{3} \frac{(1 - (2/3)^{100})}{1 - 2/3} \right]$.$S_{100} = 100 - 2 \left(1 - \left(\frac{2}{3}\right)^{100}\right) = 100 - 2 + 2 \left(\frac{2}{3}\right)^{100} = 98 + 2 \left(\frac{2}{3}\right)^{100}$.Since $0 Therefore,the greatest integer less than or equal to $S_{100}$ is $[S_{100}] = 98$.

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

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