# If $a$ and $b$ be the roots of the quadratic equation $x^2-6x+4=0$ then find the value of given expression.

Let $a$ and $b$ be the roots of the quadratic equation $x^2-6x+4=0$ and $P_n = a^n + b^n$ then the value of

$$\frac{P_{50}(P_{48}+P_{49})-6P_{49}^2+4P_{48}^2}{P_{48}.P_{49}}$$ Options are

$(A)$ $2$

$(B)$ $1$

$(C)$ $4$

$(D)$ $10$

Here roots are $x=3\pm \sqrt5$ but how are we going to calculate such big powers. Roots are real so we don't have advantage of De-Movier's theorm.

Could someone suggest something?

Note that $a+b=6$ and $ab=4$, by Vieta's theorem. On the other hand, from $$(a^{n-1}+b^{n-1})(a+b)=(a^n+b^n)+ab(a^{n-2}+b^{n-2})$$ we obtain the recursion $P_n-(a+b)P_{n-1}+ab P_{n-2}=0$, so that $$P_0=2,\quad P_1=6,\qquad P_n-6P_{n-1}+4P_{n-2}=0\quad(n\geq2)\ .$$ Using the "Master Theorem" it is possible to obtain an explicit representation of $P_n$ that should allow to compute the expression in question.