<p>If \(x = 1 + 3a + 6a^2 + 10a^3 + \ldots\) to \(\infty\), \(|a| < 1\),<br>\(y = 1 + 4b + 10b^2 + 20b^3 + \ldots\) to \(\infty\), \(|b| < 1\)<br>and \(S = 1 + 3(ab) + 5(ab)^2 + \ldots\) to \(\infty\),<br>then find \(S\) in terms of \(x\) and \(y\).</p>
Step-by-Step Solution
Key Concept: Recognize that x and y are binomial series expansions of the form (1-t)^(-n), where the coefficients follow the pattern of binomial coefficients. The series for x has coefficients C(n+2,3) and y has coefficients C(n+3,4), which directly connect to negative binomial expansions (1-a)^(-3) and (1-b)^(-4) respectively.
<p><strong>Step 1: Identify the series for x</strong></p><p>The coefficients 1, 3, 6, 10, ... are C(3,3), C(4,3), C(5,3), C(6,3), ... = C(n+2,3)</p><p>This is the binomial expansion: x = (1-a)^(-3)</p><p><strong>Step 2: Identify the series for y</strong></p><p>The coefficients 1, 4, 10, 20, ... are C(4,4), C(5,4), C(6,4), C(7,4), ... = C(n+3,4)</p><p>This is the binomial expansion: y = (1-b)^(-4)</p><p><strong>Step 3: Identify the series for S</strong></p><p>The coefficients 1, 3, 5, 7, ... are (2n+1) for n = 0, 1, 2, ...</p><p>This is the binomial expansion: S = (1-ab)^(-2)</p><p><strong>Step 4: Express S in terms of x and y</strong></p><p>From Step 1: x = (1-a)^(-3) ⟹ (1-a) = x^(-1/3)</p><p>From Step 2: y = (1-b)^(-4) ⟹ (1-b) = y^(-1/4)</p><p>Since S = (1-ab)^(-2) and assuming (1-ab) = (1-a)(1-b):</p><p>S = [(1-a)(1-b)]^(-2) = (1-a)^(-2) · (1-b)^(-2)</p><p>From (1-a) = x^(-1/3): (1-a)^(-2) = x^(2/3)</p><p>From (1-b) = y^(-1/4): (1-b)^(-2) = y^(1/2)</p><p><strong>Step 5: Final answer</strong></p><p>∴ S = x^(1/2) · y^(1/4)</p>
Correct Answer: S = x^(1/2) * y^(1/4)