$\mathrm{x}^{2}-\mathrm{ax}+\mathrm{b}=0$ has roots $\alpha, \beta$

Then $\alpha+\beta=\mathrm{a}, \alpha \beta=\mathrm{b}, \mathrm{Now}$

$ \mathrm{v}_{\mathrm{n}+1} =\alpha^{\mathrm{n}+1}+\beta^{\mathrm{n}+1} $ 

$=(\alpha+\beta)\left(\alpha^{\mathrm{n}}+\beta^{\mathrm{n}}\right) $ 

$-\alpha \beta^{\mathrm{n}}-\beta \mathrm{a}^{\mathrm{n}} $ 

$=(\alpha+\beta)\left(\alpha^{\mathrm{n}}+\beta^{\mathrm{n}}\right)-\alpha \beta\left(\alpha^{\mathrm{n}-1}+\beta^{\mathrm{n}-1}\right) $ 

$ \mathrm{v}_{\mathrm{n}+1} =\mathrm{av}_{\mathrm{n}}-\mathrm{b} \mathrm{v}_{\mathrm{n}-1} $