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:
$$ax^2+by^2=49$$
$$ax^2+by^2=(ax+by)^2$$
$$ax^2+by^2=a^2x^2+2abxy+b^2y^2$$
$$ax^2-a^2x^2+by^2-b^2y^2=2abxy$$
$$ax^2(1-a)+by^2(1-b)=2abxy$$

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 $a+b=21.$
