Question

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

Options

• \[n - \frac{1}{2}( 3^{- n} - 1)\]

• \[n - \frac{1}{2}(1 - 3^{- n} )\]

• \[n + \frac{1}{2}( 3^n - 1)\]

• \[n - \frac{1}{2}( 3^n - 1)\]

Answer 1

\[n - \frac{1}{2}(1 - 3^{- n} )\]

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:

\[S_n = \sum^n_{k = 1} T_k \]

\[ = \sum^n_{k = 1} \left[ 1 - \frac{1}{3^k} \right]\]

\[ = \sum^n_{k = 1} 1 - \sum^n_{k = 1} \frac{1}{3^k}\]

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

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

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

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