Sequences & Series
Sum of series
Grade 11
Question:
<p>If \(|a| < 1\) and \(|b| < 1\), then the sum of the series \(1 + (1+a)b + (1+a+a^2)b^2 + (1+a+a^2+a^3)b^3 + \cdots\) is</p>
<p>(1) \(\dfrac{1}{(1-a)(1-b)}\)</p>
<p>(2) \(\dfrac{1}{(1-a)(1-ab)}\)</p>
<p>(3) \(\dfrac{1}{(1-b)(1-ab)}\)</p>
<p>(4) \(\dfrac{1}{(1-a)(1-b)(1-ab)}\)</p>
Step-by-Step Solution
Key Concept: Use the formula for infinite geometric series sum = a/(1-r) when |r| < 1, and recognize that |a| < |b| with both having the same sign allows us to form valid geometric series with these parameters.
<p><strong>Step 1:</strong> Since |a| < |b|, both geometric series converge.</p><p><strong>Step 2:</strong> Sum of first series: S₁ = a/(1-b) = a + ab + ab² + ...</p><p><strong>Step 3:</strong> Sum of second series: S₂ = b/(1-a) = b + ba + ba² + ...</p><p><strong>Step 4:</strong> Both series are well-defined because |b| < 1 and |a| < 1 (since |a| < |b| < 1 is implied for convergence).</p><p><strong>Step 5:</strong> The condition |a| < |b| ensures both 1-a and 1-b are positive (for same-signed a,b), making both sums well-defined and positive.</p><p>∴ Answer: C</p>
Correct Answer: C