Let $\alpha,\ \beta$ be the roots of the equation $x^{2} - ax + b = 0$ and $A_{n} = \alpha^{n} + \beta^{n}$. Then, $A_{n + 1} - aA_{n} + bA_{n - 1}$ is equal to

Question

Dr. Abubacker · Accepted Answer

(c)
Here, $\alpha^{2} - a\alpha + b = 0$ and $\beta^{2} + a\beta + b = 0$
Now, $A_{n + 1} - aA_{n} + bA_{n - 1}$
$= \alpha^{n + 1} + \beta^{n + 1} - a\left( \alpha^{n} + \beta^{n} \right) + b(\alpha^{n - 1} + \beta^{n - 1})$
$= \alpha^{n - 1}\left( \alpha^{2} - a\alpha + b \right) + \beta^{n - 1}\left( \beta^{2} - a\beta + b \right)$
$= 0$

Q. Let $\alpha,\ \beta$ be the roots of the equation $x^{2} - ax + b = 0$ and $A_{n} = \alpha^{n} + \beta^{n}$. Then, $A_{n + 1} - aA_{n} + bA_{n - 1}$ is equal to

Explanation

Reviewed by Dr. Abubacker

Want more practice?