Basic Mathematics & Logarithm
Exponential Series
Grade 11
Question:
<p>The value of <br>\[e^{-1} = \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \cdots\]<br>is equal to</p>
<p>\(\frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \cdots\)</p>
<p>\(\frac{1}{e} = \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \cdots\)</p>
<p>\(1 - \frac{1}{2!} + \frac{1}{3!} - \cdots\)</p>
<p>None of these</p>
Step-by-Step Solution
Key Concept: Recognize that e^x = 1 + x + x²/2! + x³/3! + ... and substitute x = -1, then isolate the series starting from 1/2! by removing the first two terms (1 and -1).
<p><strong>Step 1:</strong> Recall the Taylor series expansion of e^x:</p><p>e^x = 1 + x + x²/2! + x³/3! + x⁴/4! + ...</p><p><strong>Step 2:</strong> Substitute x = -1:</p><p>e^(-1) = 1 + (-1) + (-1)²/2! + (-1)³/3! + (-1)⁴/4! + ...</p><p>e^(-1) = 1 - 1 + 1/2! - 1/3! + 1/4! - ...</p><p><strong>Step 3:</strong> The first two terms (1 - 1) cancel out:</p><p>e^(-1) = 1/2! - 1/3! + 1/4! - ...</p><p><strong>Step 4:</strong> The given series is exactly this expression.</p><p>∴ Answer: <strong>e^(-1) = 1/e</strong> (Option B)</p>
Correct Answer: B