The sum of the series 1+43+109+2827+… up to nterms is:

1+43+109+2827+… up to n terms=1+∑r=1n−1 3r+13r=1+∑r=1n−1 1+13r=1+(n−1)+131−13n−11−13=n+121−13n−1=n+12−12.3n−1

The sum of the series 1+43+109+2827+… up to nterms is:

1+43+109+2827+… up to n terms=1+∑r=1n−1 3r+13r=1+∑r=1n−1 1+13r=1+(n−1)+131−13n−11−13=n+121−13n−1=n+12−12.3n−1

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The sum of the series $1 + \frac{4}{3} + \frac{10}{9} + \frac{28}{27} + \dots$ up to $n$ terms is:

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$n - \frac{1}{3} + \frac{1}{{3.2}^{n - 1}} ∣$

$\frac{7}{6} n + \frac{1}{6} - \frac{1}{{3.2}^{n - 1}}$

$\frac{5}{3} n - \frac{7}{6} + \frac{1}{{2.3}^{n - 1}}$

$n + \frac{1}{2} - \frac{1}{{2.3}^{n - 1}}$

answer is D.

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Detailed Solution

$1 + \frac{4}{3} + \frac{10}{9} + \frac{28}{27} + \dots$ up to $n$ terms

$\begin{array}{l} = 1 + \sum_{r = 1}^{n - 1} \frac{3^{r} + 1}{3^{r}} = 1 + \sum_{r = 1}^{n - 1} 1 + {(\frac{1}{3})}^{r} \\ = 1 + (n - 1) + \frac{\frac{1}{3} \{1 - {(\begin{matrix} 1 \\ 3 \end{matrix})}^{n - 1}\}}{1 - \frac{1}{3}} = n + \frac{1}{2} (1 - \frac{1}{3^{n - 1}}) \\ = n + \frac{1}{2} - \frac{1}{{2.3}^{n - 1}} \end{array}$