Question

If $\alpha ,\beta$ are zeros of $x^2-3x+1=0$ then find the value of $\frac{{\alpha }^{2014}+{\beta }^{2014}+{\alpha }^{2016}+{\beta }^{2016}}{{\alpha }^{2015}+{\beta }^{2015}}$.

Answer 1

A] $\alpha ,\beta$ are roots of $x^2-3x+1=0$

$\therefore {\alpha }^2-3\alpha +1=0$ - (1) $\Rightarrow {\alpha }^2+1=3\alpha$

${\beta }^2-3\beta +1=0$ - (2) $\Rightarrow {\beta }^2+1=3\beta$

$=\frac{{\alpha }^{2014}+{\beta }^{2014}+{\alpha }^{2016}+{\beta }^{2016}}{{\alpha }^{2015}+{\beta }^{2015}}$

$=\frac{\alpha 2014(1+{\alpha }^2)+{\beta }^{2014}(1+{\beta }^2)}{{\alpha }^{2015}+{\beta }^{2015}}$

$=\frac{{\alpha }^{2014}\left(3\alpha \right)+{\beta }^{2014}\left(3\beta \right)}{{\alpha }^{2015}+{\beta }^{2015}}$

= 3