Question

The sum of the infinite series \( \frac{1}{5} \left( \frac{1}{5} - \frac{1}{7} \right) + \left( \frac{1}{5} \right)^2 \left( \frac{1}{5} - \frac{1}{7} \right)^2 - \left( \frac{1}{7} \right)^2 + \left( \frac{1}{5} \right)^3 \left( \frac{1}{5} - \frac{1}{7} \right)^3 + \dots \) is equal to

• A. \(\frac{7}{408}\)
• B. \(\frac{5}{408}\)
• C. \(\frac{7}{816}\)
• D. \(\frac{5}{816}\)

Answer 1

The Correct Option is B

Let's denote the given series as S:

$S = \frac{1}{5}\left(\frac{1}{5} - \frac{1}{7}\right) + \left(\frac{1}{5}\right)^2\left[\left(\frac{1}{5}\right)^2 - \left(\frac{1}{7}\right)^2\right] + \left(\frac{1}{5}\right)^3\left[\left(\frac{1}{5}\right)^3 - \left(\frac{1}{7}\right)^3\right] + \dots$

We can rewrite this as:

$S = \left(\frac{1}{5}\right)^2 - \left(\frac{1}{5}\right)\left(\frac{1}{7}\right) + \left(\frac{1}{5}\right)^4 - \left(\frac{1}{5}\right)^2\left(\frac{1}{7}\right)^2 + \left(\frac{1}{5}\right)^6 - \left(\frac{1}{5}\right)^3\left(\frac{1}{7}\right)^3 + \dots$

Now, let's group the terms:

$S = \left[\left(\frac{1}{5}\right)^2 + \left(\frac{1}{5}\right)^4 + \left(\frac{1}{5}\right)^6 + \dots\right] - \left[\left(\frac{1}{5}\right)\left(\frac{1}{7}\right) + \left(\frac{1}{5}\right)^2\left(\frac{1}{7}\right)^2 + \left(\frac{1}{5}\right)^3\left(\frac{1}{7}\right)^3 + \dots\right]$

The first series is a geometric series with first term $a = \left(\frac{1}{5}\right)^2$ and common ratio $r = \left(\frac{1}{5}\right)^2$.

The second series is also a geometric series with first term $a = \left(\frac{1}{5}\right)\left(\frac{1}{7}\right)$ and common ratio $r = \left(\frac{1}{5}\right)\left(\frac{1}{7}\right)$.

Using the formula for the sum of an infinite geometric series ($S = \frac{a}{1-r}$), we can find the sum of both series and subtract them to get the value of S.

After calculations, we get: $S = \frac{5}{408}$

Therefore, the sum of the infinite series is 5/408.