how should i go about solving the following problem?? $f(n)=a^n-b^n$ where $a$ and $b$ are roots of the following equation .$$5x^2-2x+1=0$$
Then find the value of $$\frac{5f(10)+f(9)}{f(8)}$$
I realised we can use the 5 in the equation as $\frac{1}{ab}$ or $\frac{2}{a+b}$. I proceeded with that but in vain . What should I do.
 A: This is some indirect formulation for a linear recursion. The linear recursion with
$$
5u_{n+2}-2u_{n+1}+u_n=0
$$
has exactly this polynomial as characteristic polynomial and
$$
u_n=c_1 a^n+c_2 b^n
$$
as its general solution.
Here we know that $a+b=\frac25$ and $ab=\frac15$, so that 
$$(a-b)^2=(a+b)^2-4ab=\frac4{25}-\frac45=-\frac{16}{25}.$$
Thus the initial values of the sequence are $u_0=f(0)=0$ and $u_1=f(1)=i\frac45$ (or the negative of it, resulting in the negation of all following sequence elements), the other sequence elements can now be computed via the recursion, such as
$$
u_2=\frac25u_1-\frac15u_0=i\frac8{25},\quad u_3=\frac25u_2-\frac15u_1=i\frac{16}{125}-i\frac4{25}=-i\frac{4}{125}
$$
etc.

Observing that the denominator contains increasing powers of $5$, set $v_n=5^nu_n$, then 
$$
5^{n+2}u_{n+2}-2⋅5^{n+1} u_{n+1}+5⋅5^n⋅u_n=0 \iff v_{n+2}-2⋅v_{n+1}+5⋅v_n=0,
$$
so if $v_0=0$ and $v_1$ is a (gaussian) integer, then the $v$ sequence stays in the (gaussian) integers. Using $v_1=1$ one can generate all other sequences with $u_0=0$ per $u_k=u_1⋅5^{1-n}⋅v_n$. For this choice
$$
v_0=0,\;v_1=1\; v_2=2,\; v_3=-1,\; v_4=-12,\;v_5=-19,\;v_6=22,\; v_7=129,\; v_8=148,...
$$

On observes especially, that $5f(10)+f(9)=3f(9)-f(8)$, however, $5f(10)+f(8)=2f(9)$, so
$$
\frac{5f(10)+f(8)}{f(9)}=2
$$
which would be the trivial result indicated in the comment by lab bhattacharjee.
