The sum of the series frac{2}{3} + frac{8}{9} | vedclass

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(D) The $n$-th term of the series is given by $T_n = \frac{3^n - 1}{3^n} = 1 - \left(\frac{1}{3}\right)^n$.The sum of $n$ terms is $S_n = \sum_{k=1}^{

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$n + \frac{1}{2}(3^{-n} - 1)$

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(D) The $n$-th term of the series is given by $T_n = \frac{3^n - 1}{3^n} = 1 - \left(\frac{1}{3}\right)^n$.The sum of $n$ terms is $S_n = \sum_{k=1}^{n} T_k = \sum_{k=1}^{n} \left(1 - \left(\frac{1}{3}\right)^k\right)$.$S_n = \sum_{k=1}^{n} 1 - \sum_{k=1}^{n} \left(\frac{1}{3}\right)^k$.$S_n = n - \left[ \frac{\frac{1}{3}(1 - (\frac{1}{3})^n)}{1 - \frac{1}{3}} \right]$.$S_n = n - \left[ \frac{\frac{1}{3}(1 - 3^{-n})}{\frac{2}{3}} \right]$.$S_n = n - \frac{1}{2}(1 - 3^{-n}) = n - \frac{1}{2} + \frac{1}{2} \cdot 3^{-n} = n + \frac{1}{2}(3^{-n} - 1)$.

The sum of the series $\frac{2}{3} + \frac{8}{9} + \frac{26}{27} + \frac{80}{81} + \dots$ to $n$ terms is:

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