<p>If the sum of the series \(\displaystyle\sum_{n=0}^{\infty} r^n\), \(|r| < 1\) is \(s\), then find the sum of the series \(\displaystyle\sum_{n=0}^{\infty} r^{2n}\).</p>
Step-by-Step Solution
Key Concept: The sum of an infinite geometric series ∑r^n = 1/(1-r) for |r|<1. Use this constraint to express s in terms of r, then substitute back to find the series sum in terms of s alone.
<p><strong>Step 1:</strong> Recognize that ∑_{n=0}^{∞} r^n = 1/(1-r) for |r|<1, and we're given this sum equals s.</p><p><strong>Step 2:</strong> From s = 1/(1-r), solve for r: s(1-r) = 1 ⟹ s - sr = 1 ⟹ r = (s-1)/s</p><p><strong>Step 3:</strong> Now find ∑_{n=0}^{∞} r^(2n) = ∑_{n=0}^{∞} (r²)^n = 1/(1-r²) using the geometric series formula with first term 1 and common ratio r².</p><p><strong>Step 4:</strong> Substitute r = (s-1)/s: 1/(1-r²) = 1/(1 - ((s-1)/s)²) = 1/(1 - (s-1)²/s²) = 1/((s² - (s-1)²)/s²) = s²/(s² - (s² - 2s + 1)) = s²/(2s - 1)</p><p>∴ Answer: \(\dfrac{s^2}{2s-1}\)</p>
Correct Answer: \(\dfrac{s^2}{2s-1}\)