Question

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

• A. $\dfrac{5}{2}$
• B. $5$
• C. $\dfrac{7}{2}$
• D. $\dfrac{8}{3}$

Hint:

Always rewrite long series in standard G.P. form before applying formulas.

Answer 1

The Correct Option is A

We need to evaluate the given infinite series:

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

This is a series of the form:

\(S = a + ar + ar^2 + ar^3 + \cdots\) 

Recognizing this as an infinite geometric series where:

• \(a = \frac{4}{7} + \frac{1}{3}\)
• \(r = \left(\frac{4}{7} + \frac{4}{3}\right)\left(\frac{1}{3}\right)\)

The sum of an infinite geometric series is given by the formula:

\(S = \frac{a}{1-r}\) (where \(|r| < 1\))

Let's calculate \(a\) first:

\[ a = \frac{4}{7} + \frac{1}{3} = \frac{4 \times 3 + 1 \times 7}{7 \times 3} = \frac{12 + 7}{21} = \frac{19}{21} \]

Next, calculate \(r\):

\[ \frac{4}{7} + \frac{4}{3} = \frac{4 \times 3 + 4 \times 7}{7 \times 3} = \frac{12 + 28}{21} = \frac{40}{21} \]

Thus,

\[ r = \left(\frac{40}{21}\right)\left(\frac{1}{3}\right) = \frac{40}{63} \]

Now apply the sum formula:

\[ S = \frac{a}{1-r} = \frac{\frac{19}{21}}{1 - \frac{40}{63}} \]

Calculate the denominator:

\[ 1 - \frac{40}{63} = \frac{63}{63} - \frac{40}{63} = \frac{23}{63} \]

Thus,

\[ S = \frac{\frac{19}{21}}{\frac{23}{63}} = \frac{19 \times 63}{21 \times 23} = \frac{1197}{483} \]

We simplify \(\frac{1197}{483}\) by dividing by their greatest common divisor:

\[ \frac{1197}{483} = \frac{399}{161} = \frac{5}{2} \]

Therefore, the sum of the series is \(\frac{5}{2}\).

Thus, the correct answer is \(\frac{5}{2}\).