A) \[-a\]
B) b
C) 0
D) \[a-b\]
Correct Answer: C
Solution :
We have, \[\alpha +\beta =a\] and\[\alpha \beta =0.\] Now, \[a{{A}_{n}}-n{{A}_{n-1}}=(\alpha +\beta )({{\alpha }^{n}}+{{\beta }^{n}})\] \[-\alpha \beta ({{\alpha }^{n-1}}+{{\beta }^{n-1}})\] \[={{\alpha }^{n+1}}+{{\alpha }^{n}}\beta +\alpha {{\beta }^{n}}+{{\beta }^{n+1}}-{{\alpha }^{n}}\beta -\alpha {{\beta }^{n}}\] \[={{\alpha }^{n+1}}+{{\beta }^{n+1}}\] \[={{A}_{n+1}}\]You need to login to perform this action.
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