The sum of the series $1+\frac{4}{3}+\frac{10}{9}+\frac{28}{27}+\ldots$ upto $n$ terms is

The sum of the series $1+\frac{4}{3}+\frac{10}{9}+\frac{28}{27}+\ldots$ upto $n$ terms is
  1. $\frac{7}{6} n+\frac{1}{6}-\frac{2}{3.2^{n-1}}$
  2. $\frac{5}{3} n-\frac{7}{6}+\frac{1}{2.3^{n-1}}$
  3. $n+\frac{1}{2}-\frac{1}{2.3^n}$
  4. $n-\frac{1}{3}-\frac{1}{3.2^{n-1}}$

Solution

Given series is $1+\frac{4}{3}+\frac{10}{9}+\frac{28}{27}+\ldots . n$ terms $=1+\left(1+\frac{1}{3}\right)+\left(1+\frac{1}{9}\right)+\left(1+\frac{1}{27}\right)+\ldots . n$ terms $=(1+1+1+\ldots+n$ terms $)$ $+\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\ldots n\right.$ terms $)$ $=n+\frac{\frac{1}{3}\left(1-\frac{1}{3^n}\right)}{1-\frac{1}{3}}=n+\frac{1}{3} \times \frac{3}{2}\left[1-3^{-n}\right]$ $=n+\frac{1}{2}\left[1-3^{-n}\right]=n+\frac{1}{2}-\frac{1}{2.3^n}$

Asked in: JEE Main 2012 (19 May Online)

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