If $ax+by=7$, . 
If 
  $$ax+by=7$$
  $$ax^2+by^2=49$$
  $$ax^3+by^3=133$$
  $$ax^4+by^4=406$$
  then find the value of
  $$2014(x+y-xy) - 100(a+b)$$

My attempt:
$$ax^2+by^2=49$$
$$ax^2+by^2=(ax+by)^2$$
$$ax^2+by^2=a^2x^2+2abxy+b^2y^2$$
$$ax^2-a^2x^2+by^2-b^2y^2=2abxy$$
$$ax^2(1-a)+by^2(1-b)=2abxy$$
 A: Let $\frac{ax}{ax+by}=\alpha$ and $\frac{by}{ax+by}=\beta$.
Thus, $$\alpha+\beta=1,$$
$$\alpha x+\beta y=7,$$
$$\alpha x^2+\beta y^2=19$$ and
$$\alpha x^3+\beta y^3=58.$$
Hence, $$(x+y)(\alpha x+\beta y)=7(x+y)$$ or
$$19+xy(\alpha+\beta)=7(x+y)$$ or
$$19+xy=7(x+y).$$
In another hand
$$(x+y)(\alpha x^2+\beta y^2)=19(x+y)$$ or
$$58+xy(\alpha x+\beta y)=19(x+y)$$ or
$$58+7xy=19(x+y).$$
From here we obtain $x+y=2.5$ and $xy=-\frac{3}{2}$ and the rest is smooth.
I got $a+b=21$ and $$2014(x+y-xy)-100(a+b)=2014\cdot4-100\cdot21=5956.$$ 
A: Multiplying the first three equations with $x+y$, we get
\begin{eqnarray*}
(ax+by)(x+y) &=& ax^2+by^2 +(a+b)xy \\
(ax^2+by^2)(x+y) &=& ax^3+by^3 +(ax+by)xy \\
(ax^3+by^3)(x+y) &=& ax^4+by^4 +(ax^2+by^2)xy
\end{eqnarray*}
or
\begin{eqnarray*}
7(x+y) &= & \;\;49 +(a+b)xy \\
49(x+y) &= &133 +\;\;7xy \\
133(x+y) &=& 406 +49xy
\end{eqnarray*}
Now you have three equations with three unknowns ($x+y$, $xy\;$ and $a+b$). You get $x+y = \frac{5}{2}$, $xy = -\frac{3}{2}$ and $a+b=21.$
A: The structure of the equations indicate that $7$, $49$, $133$, $406$, ... is a sequence that is defined as a homogeneous linear difference equation of order $2$ with the characteristic equation $\lambda^2+c_1\lambda +c_0=0$ with the roots $x$ and $y$. Therefore we have 
\begin{eqnarray*}
133 + \;\;49c_1 + \;\;7c_0 &=& 0 \\
406 + 133c_1 + 49c_0 &=& 0
\end{eqnarray*}
with $c_1 = -(x+y)$ and $c_0=xy$. From this, we easily get $x+y$ and $xy.$ Then we can find $x$ and $y$ and use the first two equations to figure out $a$ and $b$.
A: Hint $ $ Use $\ ax^{n+1}\!+by^{n+1} = (x\!+\!y)(ax^n\!+by^n) - xy(ax^{n-1}\!+by^{n-1})$ to solve for $\,xy,x\!+\!y$
Remark $ $ The recurrence is $\,(S-x)(S-y)f_n = (S^2 - (x+y)S+xy)f_n = 0\,$ where $\,Sf_n = f_{n+1}.\,$ It has solutions $\,f_n = x^n,\,y^n\,$ so also $\,f_n = ax^n + bx^n$ by linearity.
