Sequences & Series
Exponential Series
Grade 11
Question:
<p>The sum of series \(\dfrac{1}{2!} + \dfrac{1}{4!} + \dfrac{1}{6!} + \cdots\) is</p>
<p>\(\dfrac{e^2 - 1}{2}\)</p>
<p>\(\dfrac{(e-1)^2}{2e}\)</p>
<p>\(\dfrac{e^2 - 1}{2e}\)</p>
<p>\(\dfrac{e^2 - 2}{e}\)</p>
Step-by-Step Solution
Key Concept: Recognize that this is the sum of reciprocals of even factorials, which can be extracted from the exponential series e^x = Σ(x^n/n!). Use e^1 and e^(-1) to isolate even-factorial terms.
<p><strong>Step 1:</strong> Recall the exponential series: e^x = 1 + x + x²/2! + x³/3! + x⁴/4! + ...</p><p><strong>Step 2:</strong> For x = 1: e = 1 + 1/1! + 1/2! + 1/3! + 1/4! + 1/5! + 1/6! + ...</p><p><strong>Step 3:</strong> For x = -1: e^(-1) = 1 - 1/1! + 1/2! - 1/3! + 1/4! - 1/5! + 1/6! - ...</p><p><strong>Step 4:</strong> Add these two equations: e + e^(-1) = 2[1 + 1/2! + 1/4! + 1/6! + ...]</p><p><strong>Step 5:</strong> Therefore: 1/2! + 1/4! + 1/6! + ... = (e + e^(-1) - 2)/2 = (e + e^(-1))/2 - 1</p><p><strong>Step 6:</strong> This simplifies to: <strong>(e + 1/e)/2 - 1</strong> or equivalently <strong>(e² + 1)/(2e) - 1 = (e² - 2e + 1)/(2e) = (e-1)²/(2e)</strong></p><p>∴ Answer: <strong>B</strong> [typically (e + e^(-1))/2 - 1 or (e² + 1)/(2e) - 1]</p>
Correct Answer: B