Sequences & Series
Exponential Series
Grade 11
Question:
<p>The value of \(1 + \dfrac{1}{4 \cdot 2!} + \dfrac{1}{16 \cdot 4!} + \dfrac{1}{64 \cdot 6!} + \cdots\) is:</p>
<p>\(\dfrac{e+1}{2\sqrt{e}}\)</p>
<p>\(\dfrac{e-1}{\sqrt{e}}\)</p>
<p>\(\dfrac{e-1}{2\sqrt{e}}\)</p>
<p>\(\dfrac{e+1}{\sqrt{e}}\)</p>
Step-by-Step Solution
Key Concept: Recognize that the general term is 1/(4^n · (2n)!), which relates to the Taylor series expansion of cosh(x) = Σ(x^(2n)/(2n)!) evaluated at x=2.
<p><strong>Step 1:</strong> Write the general term of the series as a<sub>n</sub> = 1/(4<sup>n</sup> · (2n)!) for n = 1, 2, 3, ...</p><p><strong>Step 2:</strong> Recognize that cosh(x) = Σ<sub>n=0</sub>^∞ x<sup>2n</sup>/(2n)! = 1 + x²/2! + x⁴/4! + x⁶/6! + ...</p><p><strong>Step 3:</strong> Substitute x = 2: cosh(2) = 1 + 1/(1·2!) + 1/(4·4!) + 1/(16·6!) + ...</p><p><strong>Step 4:</strong> Notice our given series starts from n=1, so it equals cosh(2) - 1 = (e² + e⁻²)/2 - 1</p><p><strong>Step 5:</strong> Simplify: (e² + e⁻²)/2 - 1 = (e² + e⁻² - 2)/2 = (e - e⁻¹)²/2</p><p>∴ Answer: C</p>
Correct Answer: C