2015/110) If 'a' and 'b' are the roots of $x^2-3x+1=0$ then

find the value of $\frac{a^{2014} + b^{2014} + a^{2016} + b^{2016}}{ a^{2015} + b^{2015}}$

Solution

a is root of $x^2 – 3x + 1=0$

so $a^2 - 3a + 1 = 0$

or

$a^2 + 1 = 3a$

so

$\dfrac{a^{2014} + a^{2016}}{a^{2015}} = \dfrac{1+a^2}{a} = 3 \cdots(1)$

Similarly

$\dfrac{b^{2014} + b^{2016}}{b^{2015}} = \dfrac{1+b^2}{b} = 3 \cdots(2)$

using

if $\dfrac{x}{y} = \dfrac{z}{w}$ then both are $\dfrac{x+z}{y+w}$

we get

$\dfrac{a^{2014} + b^{2014} + a ^{2016} + b^{2016}}{a^{2015} + b^{2015}} =  3$