Question:

If \[1 + \frac{\sqrt{3} - \sqrt{2}}{2\sqrt{3}} + \frac{5 - 2\sqrt{6}}{18} + \frac{9\sqrt{3} - 11\sqrt{2}}{36\sqrt{3}} + \frac{49 - 20\sqrt{6}}{180} + \cdots\] up to \(\infty = 2 \left( \sqrt{\frac{b}{a}} + 1 \right) \log_e \left( \frac{a}{b} \right)\), where \(a\) and \(b\) are integers with \(\gcd(a, b) = 1\), then (11a + 18b\) is equal to _________.

Updated On: Nov 26, 2024
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Correct Answer: 76

Solution and Explanation

Given the infinite series:

\[ S = 1 + \frac{x}{2\sqrt{3}} + \frac{x^2}{18} + \frac{x^3}{36\sqrt{3}} + \frac{x^4}{180} + \dots, \] where \( x = \sqrt{3} - \sqrt{2} \).

Step 1: Expressing the Series Let:

\[ t = \frac{x}{\sqrt{3}} \quad \text{where} \quad x = \sqrt{3} - \sqrt{2}. \]

Rewriting the series:

\[ S = 1 + \frac{t}{2} + \frac{t^2}{6} + \frac{t^3}{12} + \frac{t^4}{20} + \dots. \]

Step 2: Using the Known Expansion From known series expansions, we have:

\[ S = 2 + \sum_{n=1}^\infty \frac{t^n}{n(n+1)}. \]

Using the expansion:

\[ S = 2 + \left(- \log(1 - t)\right) + 2. \]

Substituting \( t = \frac{\sqrt{3} - \sqrt{2}}{\sqrt{3}} \):

\[ S = 2 + \log\left(\frac{\sqrt{3}}{\sqrt{3} - \sqrt{2}}\right). \]

Step 3: Evaluating Constants Given:

\[ S = 2 \left(\sqrt{\frac{b}{a} + 1}\right) \log_e\left(\frac{a}{b}\right). \]

Comparing terms, we identify:

\[ a = 2, \quad b = 3. \]

Step 4: Calculating the Required Expression

\[ 11a + 18b = 11 \times 2 + 18 \times 3 = 22 + 54 = 76. \]

Therefore, the correct answer is \( 76 \).

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