Let $P_n=\alpha^n+\beta^n, n \in \mathbf{N}$. If $P_{10}=123, P_9=76$, $P_8=47$ and $P_1=1$, then the quadratic equation having roots $\frac{1}{\alpha}$ and $\frac{1}{\beta}$ is :

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Aicharya · Accepted Answer

$\begin{aligned} & \alpha^{10}+\beta^{10}=123 \ & \alpha+\beta=1 \ & \alpha^9+\beta^9=76 \ & \alpha^8+\beta^8=47 \ & P_{10}=P_9+P_8 \ & x^2=x+1 \Rightarrow x^2-x-1=0 \ & \alpha+\beta=1, \alpha \beta=-1 \ & \frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha \beta}=\frac{1}{-1}=-1, \frac{1}{\alpha \beta}=-1\end{aligned}$ ~

Let $P_n=\alpha^n+\beta^n, n \in \mathbf{N}$. If $P_{10}=123, P_9=76$, $P_8=47$ and $P_1=1$, then the…

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