QJEEM24Mab24a: JEE Main 2024 Mathematics - Class 11 JEEM24Mab24a Solution: If upto , where a and b are integers with , then is equal to…

$\begin{aligned}& S=1+\frac{\sqrt{3}-\sqrt{2}}{2 \sqrt{3}}+\frac{5-2 \sqrt{6}}{18}+\frac{9 \sqrt{3}-11 \sqrt{2}}{36 \sqrt{3}}+\ldots \infty \\\\& =1+\frac{(1-\sqrt{2} / \sqrt{3})}{2}+\frac{(1-\sqrt{2} / \sqrt{3})^2}{6}+\frac{(1-\sqrt{2} / \sqrt{3})^3}{12}+\ldots \infty\end{aligned}$

$\text { let } 1-\frac{\sqrt{2}}{\sqrt{3}}=a$

$\begin{aligned}& S=1+\frac{a}{2}+\frac{a^2}{6}+\frac{a^3}{12}+\ldots \\\\& =1+\left(1-\frac{1}{2}\right) a+\left(\frac{1}{2}-\frac{1}{3}\right) a^2+\left(\frac{1}{3}-\frac{1}{4}\right) a^3+\ldots \\\\& =1+\left(a+\frac{a^2}{2}+\frac{a^3}{3} \ldots \infty\right)+\frac{1}{a}\left(\frac{-a^2}{2}-\frac{a^3}{3}-\frac{a^4}{4} \ldots \infty\right) \\\\& =-\ln (1-a)+\frac{1}{a}\left(-a-\frac{a^2}{2}-\frac{a^3}{3} \ldots \infty\right)+2 \\\\& =-\ln (1-a)+\frac{1}{a} \ln (1-a)+2 \\\\& =2+\left(\frac{1}{a}-1\right) \ln (1-a) \\\\& =2+\left(\frac{\sqrt{3}}{\sqrt{3}-\sqrt{2}}-1\right) \ln \left(1-1+\sqrt{\frac{2}{3}}\right) \\\\& =2+\frac{\sqrt{2}}{\sqrt{3}}-\sqrt{2} \ln \sqrt{\frac{2}{3}} \\\\& =2+\left(\frac{\sqrt{6}+2}{1} \cdot \frac{1}{2} \ln \frac{2}{3}\right) \\\\& \therefore 2+\left(\sqrt{\frac{3}{2}}+1\right) \ln \frac{2}{3} \\\\& \therefore 11 a+18 b=76\end{aligned}$