10
$\begingroup$

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?

$\endgroup$
4
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.