Question

The value of
\[ \left(\frac{1}{3}+\frac{4}{7}\right) +\left(\frac{1}{3^2}+\frac{1}{3}\times\frac{4}{7}+\frac{4}{7^2}\right) +\left(\frac{1}{3^3}+\frac{1}{3^2}\times\frac{4}{7}+\frac{1}{3}\times\frac{4}{7^2}+\frac{4}{7^3}\right) +\cdots \text{ up to infinite terms is} \]

• A. $\dfrac{7}{4}$
• B. $\dfrac{4}{3}$
• C. $\dfrac{6}{5}$
• D. $\dfrac{5}{2}$

Hint:

When terms grow in layered powers, try rewriting the expression as a geometric series.

Answer 1

The Correct Option is B

To find the sum of the given series:

\[ \left(\frac{1}{3}+\frac{4}{7}\right) +\left(\frac{1}{3^2}+\frac{1}{3}\times\frac{4}{7}+\frac{4}{7^2}\right) +\left(\frac{1}{3^3}+\frac{1}{3^2}\times\frac{4}{7}+\frac{1}{3}\times\frac{4}{7^2}+\frac{4}{7^3}\right) +\cdots \]

Let's analyze the series step-by-step:

1. The series is structured as a sum of infinite terms, each being a geometric series. The general term of the series is \[ \left(\frac{1}{3^n} + \frac{1}{3^{n-1}}\times\frac{4}{7} + \frac{1}{3^{n-2}}\times\frac{4^2}{7^2} + \ldots + \frac{4^n}{7^n}\right) \] for n starting from 1 to infinity. Each sub-term in the general term is a specific geometric progression term.
2. For each of these segments which are also geometric progressions, we identify the first term and the common ratio.
3. If we consider the overall series, it can be split into geometric series with the general form: \[ S = \sum_{n=1}^{\infty}\left( \frac{4}{21} \right)^n \] where \frac{4}{21} is the common ratio of the primary geometric series.
4. The formula for the sum S of an infinite geometric series is \[ S = \frac{a}{1 - r} \] where a is the first term and r is the common ratio.
5. Here, the first term a is the sum of the first part of the series: \[ S = \frac{\frac{7}{21}}{1 - \frac{4}{21}} \]
6. On simplification: \[ S = \frac{\frac{1}{3}}{\frac{17}{21}} = \frac{7}{17} \times \frac{21}{1} = \frac{7 \times 21}{17 \times 3} = \frac{147}{51} = \frac{4}{3} \]

Thus, the correct value of the series is \(\frac{4}{3}\).

The correct answer is: 4/3