Question

The sum of the series $\frac{2}{3}+\frac{8}{9}+\frac{26}{27}+\frac{80}{81}+$..... to n terms is
(a) $n-\frac{1}{2}(3^{-n}-1)$

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

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

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

Answer 1

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

Let $T_n$ be the nth 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:

$$\begin{gathered} S_n=\sum _{k=1}^n{T}_k \\ =\sum _{k=1}^n\left[1-\frac{1}{3^k}\right] \\ =\sum _{k=1}^{n}1-\sum _{k=1}^n\frac{1}{3^k} \\ =n-\left[\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^n}\right] \\ =n-\frac{1}{3}\left[\frac{1-{\left({\displaystyle \frac{1}{3}}\right)}^n}{1-{\displaystyle \frac{1}{3}}}\right] \\ =n-\frac{1}{2}\left[1-{\left(\frac{1}{3}\right)}^n\right] \\ =n-\frac{1}{2}\left[1-3^{-n}\right] \end{gathered}$$