Question

If \( x \) is positive then the sum to infinity of the series \[ \frac{1}{1+3x} - \frac{1}{1+3x^2} + \frac{1}{1+3x^3} - \dots \, \infty \] is:

• A. \( \frac{1}{6x(1+3x)} \)
• B. \( \frac{1}{6x} \)
• C. \( \frac{1}{2(1+3x)} \)
• D. \( \frac{1}{(1+3x)^3} \)

Hint:

Use the formula for the sum of an infinite geometric series to find the sum of such series.

Answer 1

The Correct Option is A

Step 1: Use the formula for sum of infinite series.
Use the formula for sum of infinite geometric series to calculate the sum. After simplifying, we get \( \frac{1}{6x(1+3x)} \).
Thus, the correct answer is (1).