A) \[-a\]
B) \[b\]
C) 0
D) \[a-b\]
Correct Answer: C
Solution :
\[\because \]\[\alpha \] and \[\beta \] be the roots of\[{{x}^{2}}-ax+b=0\]. \[\therefore \]\[{{\alpha }^{2}}-a\alpha +b=0\] and \[{{\beta }^{2}}-\alpha \beta +b=0\] Now, \[{{A}_{n+1}}-a{{A}_{n}}+b{{A}_{n-1}}\] \[={{\alpha }^{n+1}}+{{\beta }^{n+1}}-\alpha ({{\alpha }^{n}}+{{\beta }^{n}})+b({{\alpha }^{n-1}}+{{\beta }^{n-1}})\] \[={{\alpha }^{n-1}}({{\alpha }^{2}}-a\alpha +b)+{{\beta }^{n-1}}({{\beta }^{2}}-a\beta +b)\] \[=0\]
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