$\begin{array}{l}Let\alpha and\beta betworealnumberssuchthat\alpha +\beta =1and\alpha \beta =-1let{P}_n={\alpha }^n+{\beta }^n,P_{n-1}=11and{P}_{n+1}=29\\ persomeitegern\ge 1thenthevalueof{P}_n^2=\end{array}$

$\begin{array}{l}Given\alpha +\beta =1,\alpha \beta =-1\\ \therefore Equationis{x}^2-x-1=0\\ \alpha ,\beta arerootsof{x}^2-x-1=0\\ \Rightarrow {\alpha }^2-\alpha -1=0\Rightarrow {\alpha }^{n+1}={\alpha }^n+{\alpha }^{n-1}--\left(1\right)\\ and{\beta }^2-\beta -1=0\Rightarrow {\beta }^{n+}={\beta }^n+{\beta }^{n-1}--\left(2\right)\\ \left(1\right)+\left(2\right)\\ {\alpha }^{n+1}+{\beta }^{n+1}=({\alpha }^n+{\beta }^n)+({\alpha }^{n-1}+{\beta }^{n-1})\\ \Rightarrow P_{n+1}=P_n+P_{n-1}\\ \Rightarrow 29=P_n+11\\ \Rightarrow P_n=18\\ \therefore P_n^2=324\end{array}$