Question

If \( x = 1 + 2a + 3a^2 + 4a^3 + \ldots \) (\(|a| < 1\)) and \( y = 1 + 3b + 6b^2 + \ldots\) (\(|b| < 1\)), then find \( 1 + ab + (ab)^2 + (ab)^3 + \ldots \) in terms of \(x\) and \(y\).

• A. \( \frac{x^{1/2}y^{1/3}}{x^{1/2} + y^{1/3} - 1} \)
• B. \( \frac{xy}{x + y - 1} \)
• C. \( \frac{x^{1/3}y^{2/3}}{x^{1/3}y^{1/2} - 1} \)
• D. None of these

Hint:

Transform complex series into GP or derivative-of-GP forms, then back-substitute to get neat expressions.

Answer 1

The Correct Option is A

We identify \(x\) and \(y\) as sums related to derivatives of GP series: \[ x = \frac{1}{(1-a)^2}, \quad y = \frac{1}{(1-b)^3} \ \text{(after algebraic manipulation)} \] From these, solve for \(a\) and \(b\), then find \(ab\). We have: \[ 1 + ab + (ab)^2 + \ldots = \frac{1}{1 - ab} \] Substituting \(a\) and \(b\) in terms of \(x\) and \(y\), we get: \[ \frac{x^{1/2}y^{1/3}}{x^{1/2} + y^{1/3} - 1} \]