Sequences & Series
Exponential Series
Grade 11
Question:
<p>The value of \(\dfrac{1}{2!} + \dfrac{1}{4!} + \dfrac{1}{6!} + \cdots\) is:</p>
<p>\(\dfrac{e+e^{-1}}{2}\)</p>
<p>\(\dfrac{(e-1)^2}{2e}\)</p>
<p>\(\dfrac{e^2-1}{2e}\)</p>
<p>\(\dfrac{e-e^{-1}}{2}\)</p>
Step-by-Step Solution
Key Concept: Recognize that this sum is related to the Taylor series for e^x and e^(-x). By adding e^x and e^(-x), the odd-powered terms cancel, leaving only even-powered terms, which can be isolated.
<p><strong>Step 1:</strong> Recall the Taylor series: $e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} = 1 + \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots$</p><p><strong>Step 2:</strong> Write out e¹ and e⁻¹:<br/>$e = 1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + \cdots$<br/>$e^{-1} = 1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \cdots$</p><p><strong>Step 3:</strong> Add these two equations:<br/>$e + e^{-1} = 2\left(1 + \frac{1}{2!} + \frac{1}{4!} + \frac{1}{6!} + \cdots\right)$<br/>(Odd terms cancel, even terms double)</p><p><strong>Step 4:</strong> Solve for the desired sum:<br/>$\frac{1}{2!} + \frac{1}{4!} + \frac{1}{6!} + \cdots = \frac{e + e^{-1}}{2} - 1 = \frac{e + e^{-1} - 2}{2}$</p><p>Or equivalently: $\frac{e + e^{-1}}{2} - 1$</p><p>∴ Answer: <strong>B</strong> (which is $\frac{e + e^{-1}}{2} - 1$ or $\frac{e + e^{-1} - 2}{2}$)</p>
Correct Answer: B