If 1+frac{sqrt{3}-sqrt{2}}{2 sqrt{3}}+frac{5- | vedclass

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(A) Let the given series be $S = 1 + \frac{x}{2 \sqrt{3}} + \frac{x^2}{18} + \frac{x^3}{36 \sqrt{3}} + \frac{x^4}{180} + \ldots \infty$,where $x = \sq

Vedclass · Accepted Answer

$76$

Vedclass · Answer

(A) Let the given series be $S = 1 + \frac{x}{2 \sqrt{3}} + \frac{x^2}{18} + \frac{x^3}{36 \sqrt{3}} + \frac{x^4}{180} + \ldots \infty$,where $x = \sqrt{3} - \sqrt{2}$.Substitute $t = \frac{x}{\sqrt{3}} = \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}} = 1 - \sqrt{\frac{2}{3}}$.The series becomes $S = 1 + \frac{t}{2} + \frac{t^2}{6} + \frac{t^3}{12} + \frac{t^4}{20} + \ldots = 1 + \sum_{n=1}^{\infty} \frac{t^n}{n(n+1)}$.Using partial fractions,$\frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}$.$S = 1 + \sum_{n=1}^{\infty} \left(\frac{1}{n} - \frac{1}{n+1}\right) t^n = 1 + \sum_{n=1}^{\infty} \frac{t^n}{n} - \sum_{n=1}^{\infty} \frac{t^n}{n+1}$.We know $-\log_e(1-t) = \sum_{n=1}^{\infty} \frac{t^n}{n}$.Also,$\sum_{n=1}^{\infty} \frac{t^n}{n+1} = \frac{1}{t} \sum_{n=1}^{\infty} \frac{t^{n+1}}{n+1} = \frac{1}{t} \left(-\log_e(1-t) - t\right)$.Thus,$S = 1 - \log_e(1-t) - \frac{1}{t} (-\log_e(1-t) - t) = 1 - \log_e(1-t) + \frac{1}{t} \log_e(1-t) + 1 = 2 + \left(\frac{1}{t} - 1\right) \log_e(1-t)$.Since $1-t = \sqrt{\frac{2}{3}}$,we have $S = 2 + \left(\frac{1}{1-\sqrt{2/3}} - 1\right) \log_e\left(\sqrt{\frac{2}{3}}\right) = 2 + \left(\frac{\sqrt{3}}{\sqrt{3}-\sqrt{2}} - 1\right) \log_e\left(\sqrt{\frac{2}{3}}\right)$.$S = 2 + \left(\frac{\sqrt{2}}{\sqrt{3}-\sqrt{2}}\right) \log_e\left(\sqrt{\frac{2}{3}}\right) = 2 + \sqrt{2}(\sqrt{3}+\sqrt{2}) \cdot \frac{1}{2} \log_e\left(\frac{2}{3}\right) = 2 + (\sqrt{6}+2) \cdot \frac{1}{2} \log_e\left(\frac{2}{3}\right) = 2 + \left(\sqrt{\frac{6}{4}} + 1\right) \log_e\left(\frac{2}{3}\right) = 2 + \left(\sqrt{\frac{3}{2}} + 1\right) \log_e\left(\frac{2}{3}\right)$.Comparing with $2\left(\sqrt{\frac{b}{a}}+1\right) \log_e\left(\frac{a}{b}\right)$,we get $a=2, b=3$.$11a + 18b = 11(2) + 18(3) = 22 + 54 = 76$.

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