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: $$ax^2+by^2=49$$ $$ax^2+by^2=(ax+by)^2$$ $$ax^2+by^2=a^2x^2+2abxy+b^2y^2$$ $$ax^2-a^2x^2+by^2-b^2y^2=2abxy$$ $$ax^2(1-a)+by^2(1-b)=2abxy$$
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 $$2014(x+y-xy)-100(a+b)=2014\cdot4-100\cdot21=5956.$$
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 $a+b=21.$
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 $b$.
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.
