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Let roots of equation $x^2-ax+b$ be $\alpha,\beta$.Given $A_n=\alpha^n+\beta^n$ then,$A_{n+1}-aA_n+bA_{n-1}$ is equal to

$\begin{array}{1 1}(A)\;-a&(B)\;0\\(C)\;b&(D)\;(a-b)\end{array} $

1 Answer

$A_{n+1}-aA_n+bA_{n-1}=\alpha^{n+1}+\beta^{n+1}-a(\alpha^n+\beta^n)+b(\alpha^{n-1}+\beta^{n-1})$
$\Rightarrow \alpha^{n+1}+\beta^{n+1}+\alpha^{n-1}(b-a\alpha)+\beta^{n-1}(b-a\beta)$
$\Rightarrow (\alpha^{n+1}+\beta^{n+1})+\alpha^{n-1}(\alpha\beta-(\alpha+\beta)\alpha)+\beta^{n-1}(\alpha\beta-(\alpha+\beta)\beta)$
$\Rightarrow (\alpha^{n+1}+\beta^{n+1})-(\alpha^{n+1}+\beta^{n+1})$
$\Rightarrow 0$
Hence (B) is the correct answer.
answered Apr 15, 2014 by sreemathi.v
 

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