Math Olympiad Algebra Question If,

$$ax + by = 7$$ $$ax^2 + by^2 = 49$$ $$ax^3 + by^3 = 133$$ $$ax^4 + by^4 = 406$$

Then find the value of$-$

$$2014(x+y-xy) - 100(a+b)$$

I came across this question in a Math Olympiad Competition and I am not sure how to solve it. Can anyone help? Thanks.
 A: These equations are in the form of the general solution of a linear recurrence equation of order $2$ with constant coefficients. The general formula of the sequence is
$$
u_n=ax^n+by^n
$$
as the solution of a recurrence
$$
u_{n+1}=cu_n+du_{n-1},
$$
where $x,y$ are solutions of the characteristic equation
$$
0=λ^2-cλ-d=(λ-x)(λ-y).
$$
Knowing two triples from the given four members $(u_1,u_2,u_3,u_4)=(7,49,133,406)=7\cdot(1,7,19,58)$ of the sequence allows to establish two linear equations for $c$ and $d$
\begin{align}
19&=7c+d\\
58&=19c+7d\\
\text{and consequently}&\\
75&=30c\\ c&=\frac52\\
d&=19-7c=\frac32
\end{align}
so that
$$
2u_{n+2}-5u_{n+1}-3u_n=0.
$$
For the roots $x$ and $y$ of the characteristic equation we get $x+y=\frac52$ and $xy=-\frac32$. The expression $a+b=u_0$ is the term of the sequence before the given ones,
$$
a+b=\frac13(2u_2-5u_1)=\frac13(98-35)=21
$$
This is sufficient to compute the crazy combination that is requested as answer.
A: From $ax+by=7$, we have $ax=7-by, by=7-ax$.
Noting
$$ ax^2+by^2=x\cdot ax+y\cdot by=x(7-by)+y(7-ax)=7(x+y)-(a+b)xy, $$
from $ax^2+by^2=49$, we obtain
$$ \tag{$*$} 7(x+y)-(a+b)xy=49. $$
Similarly, 
$$ ax^2=49-by^2,by^2=49-ax^2, ax^3=133-by^3,by^3=133-ax^3, $$
from which, we have
$$ ax^3+by^3=x\cdot ax^2+y\cdot by^2=x(49-by^2)+y(49-ax^2)=49(x+y)-(ax+by)xy $$
and hence $49(x+y)-7xy=133$
or
$$ \tag{$**$} 7(x+y)-xy=19. $$
Finally,
$$ ax^4+by^4=x\cdot ax^3+y\cdot by^3=x(133-by^3)+y(133-ax^3)=133(x+y)-(ax^2+by^2)xy=133(x+y)-49xy $$
and hence $133(x+y)-49xy=406$ or 
$$\tag{$***$}\quad\quad\quad 19(x+y)-7xy=58. $$
From $(*), (**), (***)$, it is easy to see
$$x+y=\frac{5}{2},xy=-\frac{3}{2},a+b=21 $$
and hence
$$ 2014(x+y-xy)-100(a+b)=5956. $$
