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.
Best Answer
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.