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JEE Main 2024
05-04-2024 S2
Question
If $1+\frac{\sqrt{3}-\sqrt{2}}{2 \sqrt{3}}+\frac{5-2 \sqrt{6}}{18}+\frac{9 \sqrt{3}-11 \sqrt{2}}{36 \sqrt{3}}+\frac{49-20 \sqrt{6}}{180}+\cdots$. upto $\infty=2\left(\sqrt{\frac{b}{a}}+1\right) \log _e\left(\frac{a}{b}\right)$, where a and b are integers with $\operatorname{gcd}(a, b)=1$, then $11 a+18 b$ is equal to $\_\_\_\_$ .
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Solution
$\begin{aligned} & S=1+\frac{x}{2 \sqrt{3}}+\frac{x^2}{18}+\frac{x^3}{36 \sqrt{3}}+\frac{x^4}{180}+\cdots \infty \\ & \text { Put } \frac{x}{\sqrt{3}}=t, \text { where } x=\sqrt{3}-\sqrt{2} \\ & S=1+\frac{t}{2}+\frac{t^2}{6}+\frac{t^3}{12}+\frac{t^4}{20}+\cdots \\ & S=1+t\left(1-\frac{1}{2}\right)+t^2\left(\frac{1}{2}-\frac{1}{3}\right)+t^3\left(\frac{1}{3}-\frac{1}{4}\right)+t^4\left(\frac{1}{4}-\frac{1}{5}\right) \\ & S=\left(1+t+\frac{t^2}{2}+\frac{t^3}{3}+\frac{t^3}{4}+\cdots\right)-\left(\frac{t}{2}+\frac{t^2}{3}+\frac{t^3}{4}+\frac{t^4}{5}+\cdots\right) \\ & S=\left(t+\frac{t^2}{2}+\cdots\right)-\frac{1}{t}\left(t+\frac{t^2}{2}+\frac{t^3}{3}+\cdots\right)+2\end{aligned}$
$\begin{aligned} & S=2+\left(1-\frac{1}{t}\right)(-\log (1-t))=\left(\frac{1}{t}-1\right) \log (1-t)+2 \\ & S=2+\left(\frac{\sqrt{3}}{\sqrt{3}-\sqrt{2}}-1\right) \log \left(1-\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}}\right) \\ & S=2+\left(\frac{\sqrt{2}}{\sqrt{3}-\sqrt{2}}\right) \log e \frac{\sqrt{2}}{\sqrt{3}} \\ & S=2+\frac{(\sqrt{6}+2)}{2} \log e \frac{2}{3}=2+\left(\sqrt{\frac{3}{2}}+1\right) \log e \frac{2}{3} \\ & a=2, b=3 \\ & 11 a+18 b=11 \times 2+18 \times 3=76\end{aligned}$
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