Question

The sum of n terms of the series $ \frac{4}{3}+\frac{10}{9}+\frac{28}{27}+... $ is:

• A. $ \frac{3^{n}(2n+1)+1}{2(3^{n})} $
• B. $ \frac{3^{n}(2n+1)-1}{2(3^{n})} $
• C. $ \frac{3^{n}n-1}{2(3^{n})} $
• D. $ \frac{3^{n}-1}{2} $

Answer 1

Answer: (B)

The given series can be rewritten as
$ \left( 1+\frac{1}{3} \right)+\left( 1+\frac{1}{9} \right)+\left( 1+\frac{1}{27} \right)+...n $
$ =n+\frac{1}{3}\left( 1+\frac{1}{3}+\frac{1}{3^{2}}+...n\,\,\text{terms} \right) $
$ =n+\frac{1}{3}\frac{\left[ 1-{{\left( \frac{1}{3} \right)}^{n}} \right]}{1-\frac{1}{3}}$
$=n+\frac{\frac{1}{3}\left( 1-\frac{1}{{{3}^{n}}} \right)}{\frac{2}{3}} $
$ =\frac{2n+1-\frac{1}{3^{n}}}{2}=\frac{3^{n}(2n+1)-1}{2\cdot 3^{n}} $