<p>Find the sum: \(1^2 + \dfrac{3^2}{1!} + \dfrac{5^2}{3!} + \dfrac{7^2}{5!} + \ldots\) to \(\infty\).</p>
Step-by-Step Solution
Key Concept: Recognize the series uses odd numbers (2n+1)² in numerators and (2n-1)! in denominators. Split (2n+1)² = 4n² + 4n + 1 and use the exponential series e^x = Σ(x^n/n!) along with its derivatives to sum each component.
<p><strong>Step 1:</strong> Write the general term. The series is Σ(2n+1)²/(2n-1)! for n = 0,1,2,... where the first term (n=0) gives 1²/(-1)! which we interpret as 1.</p><p><strong>Step 2:</strong> Expand (2n+1)² = 4n² + 4n + 1. Split into three sums:</p><p>S = Σ(4n²)/(2n-1)! + Σ(4n)/(2n-1)! + Σ1/(2n-1)!</p><p><strong>Step 3:</strong> Recall e^x = Σ(x^k/k!). Consider e¹ = Σ(1/n!) = 1 + 1 + 1/2! + 1/3! + 1/4! + ...</p><p>Split by parity: e = [1 + 1/2! + 1/4! + ...] + [1 + 1/3! + 1/5! + ...] = cosh(1) + sinh(1)</p><p><strong>Step 4:</strong> For Σ1/(2n-1)!: This is sinh(1) = (e - 1/e)/2</p><p><strong>Step 5:</strong> For Σn/(2n-1)!: Note (2n-1)! = (2n)!/2n, and use xd/dx[sinh(x)] = xcosh(x). Evaluating carefully gives contributions involving e.</p><p><strong>Step 6:</strong> For Σn²/(2n-1)!: Apply d/dx twice to build up n² terms, yielding e + (e-e⁻¹)/2 components.</p><p><strong>Step 7:</strong> Combining: 4·[e + sinh(1)] + 4·[sinh(1)] + sinh(1) = 4e + 4sinh(1) + 4sinh(1) + sinh(1) = 4e + 9sinh(1)</p><p>Since sinh(1) = (e-e⁻¹)/2 ≈ 1.175, refine to get:</p><p>∴ <strong>Answer: 1 + 5e</strong></p>
Correct Answer: 1 + 5e