QJEEM24Ma6afd: JEE Main 2024 Mathematics - Class 11 JEEM24Ma6afd Solution: Let , be the roots of the equation . Let . Then is equal to

$\begin{aligned}& x^2-\sqrt{2} x-\sqrt{3}=0 \\& P_n=\alpha^n-\beta^n\end{aligned}$

$\alpha$ and $\beta$ are the roots of the equation

Using Newton's theorem

$\begin{aligned}& P_{n+2}-\sqrt{2} P_{n+1}-\sqrt{3} P_n=0 \\& \text { Put } n=10 \\& P_{12}-\sqrt{2} P_{11}-\sqrt{3} P_{10}=0\end{aligned}$

$P_{12}=\sqrt{2} P_{11}+\sqrt{3} P_{10}$

Put $n=9$

$\begin{aligned}& P_{11}-\sqrt{2} P_{10}-\sqrt{3} P_9=0 \\& P_{11}=\sqrt{2} P_{10}+\sqrt{3} P_9 \\& (11 \sqrt{3}-10 \sqrt{2}) P_{10}+(11 \sqrt{2}+10) P_{11}-11 P_{12}\end{aligned}$

Put the value of $P_{12}$ & $P_{12}$ in above equation.

$\begin{aligned}= & (11 \sqrt{3}-10 \sqrt{2}) P_{10}+(11 \sqrt{2}+10)\left(\sqrt{2} P_{10}+\sqrt{3} P_9\right)-11\left(\sqrt{2} P_{11}+\sqrt{3} P_{10}\right) \\= & 11 \sqrt{3} P_{10}-10 \sqrt{2} P_{10}+22 P_{10}+10 \sqrt{2} P_{10}+11 \sqrt{6} P_9+10 \sqrt{3} P_9-11 \sqrt{2} P_{11}-11 \sqrt{3} P_{10} \\= & 22 P_{10}+11 \sqrt{6} P_9+10 \sqrt{3} P_9-11 \sqrt{2}\left(\sqrt{2} P_{10}+\sqrt{3} P_9\right) \\= & 22 P_{10}+11 \sqrt{6} P_9+10 \sqrt{3} P_9-22 P_{10}-11 \sqrt{6} P_9 \\= & 10 \sqrt{3} P_9\end{aligned}$