Q94 / 180 · COMEDK 2023 (Memory Based) · Mathematics
The sum of $n$ terms of the series, $\frac{4}{3}+\frac{10}{9}+\frac{28}{27}+\ldots$ is
- A$\frac{3^n(2 n+1)+1}{2\left(3^n\right)}$
- Correct answerB$\frac{3^n(2 n+1)-1}{2\left(3^n\right)}$
- C$\frac{3^n n-1}{2\left(3^n\right)}$
- D$\frac{3^n-1}{2}$
Solution
Given series is
$\frac{4}{3}+\frac{10}{9}+\frac{28}{27}+\ldots$
The sum of the given series upto $n$-terms
$\begin{aligned}
& \quad \frac{4}{3}+\frac{10}{9}+\frac{28}{27}+\ldots . \text { upto } n \text {-terms } \\
& =\left(1+\frac{1}{3}\right)+\left(1+\frac{1}{9}\right)+\left(1+\frac{1}{27}\right)+\ldots \text { upto } n \text {-terms } \\
& =(1+1+1+\ldots \text { upto } n \text {-terms }) \\
& +\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\ldots \text { upto } n \text { - terms }\right) \\
& =n+\frac{1}{3}\left(\frac{1-\frac{1}{3^n}}{1-\frac{1}{3}}\right)=n+\frac{\left(3^n-1\right)}{2\left(3^n\right)}=\frac{3^n(2 n+1)-1}{2\left(3^n\right)}
\end{aligned}$
Hence, the correct answer is option: 'B'.