Sequences & Series
Exponential Series
Grade 11
Question:
<p>The sum of the series \(1 + \dfrac{1}{4 \cdot 2!} + \dfrac{1}{16 \cdot 4!} + \dfrac{1}{64 \cdot 6!} + \cdots \infty\) is</p>
<p>\(\dfrac{e-1}{\sqrt{e}}\)</p>
<p>\(\dfrac{e+1}{\sqrt{e}}\)</p>
<p>\(\dfrac{e-1}{2\sqrt{e}}\)</p>
<p>\(\dfrac{e+1}{2\sqrt{e}}\)</p>
Step-by-Step Solution
Key Concept: Recognize this as a series of the form Σ(1/(4^n·(2n)!)) and connect it to the Taylor expansion of hyperbolic functions, specifically cosh(x) = Σ(x^(2n)/(2n)!) or sinh(x)/x. Setting x = 1 in cosh(1) - 1 + appropriate terms yields the answer.
<p><strong>Step 1:</strong> Identify the general term of the series.</p><p>The series is: 1 + 1/(4·2!) + 1/(16·4!) + 1/(64·6!) + ...</p><p>General term: a_n = 1/(4^n·(2n)!) for n = 0, 1, 2, ...</p><p><strong>Step 2:</strong> Rewrite using known Taylor expansions.</p><p>Recall: e^x = Σ(x^n/n!) and cosh(x) = Σ(x^(2n)/(2n)!) = (e^x + e^(-x))/2</p><p>Our series: Σ_{n=0}^∞ (1/2^(2n))·1/(2n)! = Σ_{n=0}^∞ ((1/2)^(2n))/(2n)!</p><p><strong>Step 3:</strong> Recognize this matches cosh(1/2).</p><p>Since cosh(x) = Σ_{n=0}^∞ (x^(2n))/(2n)!, we have:</p><p>cosh(1/2) = Σ_{n=0}^∞ ((1/2)^(2n))/(2n)! = (e^(1/2) + e^(-1/2))/2</p><p><strong>Step 4:</strong> Calculate the final answer.</p><p>cosh(1/2) = (√e + 1/√e)/2 = (e + 1)/(2√e) = <strong>(e + 1)/(2√e)</strong> or equivalently <strong>√e/2 + 1/(2√e)</strong></p><p>∴ Answer: D</p>
Correct Answer: D