The sum of the series

Question:

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

(a) $n-\frac{1}{2}\left(3^{-n}-1\right)$

(b) $n-\frac{1}{2}\left(1-3^{-n}\right)$

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

(d) $n-\frac{1}{2}\left(3^{n}-1\right)$

Solution:

(b) $n-\frac{1}{2}\left(1-3^{-n}\right)$

Let $T_{n}$ be the $n$th term of the given series.

Thus, we have:

$T_{n}=\frac{3^{n}-1}{3^{n}}=1-\frac{1}{3^{n}}$

Now,

Let $S_{n}$ be the sum of $n$ terms of the given series.

Thus, we have:

$S_{n}=\sum_{k=1}^{n} T_{k}$

$=\sum_{k=1}^{n}\left[1-\frac{1}{3^{k}}\right]$

$=\sum_{k=1}^{n} 1-\sum_{k=1}^{n} \frac{1}{3^{k}}$

$=n-\left[\frac{1}{3}+\frac{1}{3^{2}}+\frac{1}{3^{3}}+\ldots+\frac{1}{3^{n}}\right]$

$=n-\frac{1}{3}\left[\frac{1-\left(\frac{1}{3}\right)^{n}}{1-\frac{1}{3}}\right]$

$=n-\frac{1}{2}\left[1-\left(\frac{1}{3}\right)^{n}\right]$

$=n-\frac{1}{2}\left[1-3^{-n}\right]$

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