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$f(n)=a^n-b^n$ 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.

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  • $\begingroup$ $f$ is complex-valued -- is it on purpose? (the roots are not real). $\endgroup$ – Clement C. May 11 '14 at 14:25
  • $\begingroup$ Related :math.stackexchange.com/questions/789901/… $\endgroup$ – lab bhattacharjee May 11 '14 at 14:35
  • $\begingroup$ @user142634, Can you please cross check the Question. I think it should be $$\frac{5f(10)+f(8)}{f(9)}$$ $\endgroup$ – lab bhattacharjee May 11 '14 at 14:37
  • $\begingroup$ Even i thought so but it is f(9) in the numerator $\endgroup$ – avz2611 May 11 '14 at 14:37
  • $\begingroup$ How about getting the roots with the standard quadratic formula and then slogging through the arithmetic to calculate the result ? $\endgroup$ – Tom Collinge May 11 '14 at 15:23
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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.

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