Question

Assertion :Let $\alpha ,\beta$ be the roots of the equation $x^2-ax+b=0$. If the coordinates of $A_n$ are $({\alpha }^{n/2},{\beta }^{n/2})$ then,$(O{A}_{n+1}{)}^2-a(O{A}_n{)}^2+b(O{A}_{n-1}{)}^2$ is equal to zero,O being the origin Reason: If $\alpha ,\beta$ are the roots of the equation $x^2-ax+b=0$, then ${\alpha }^n+{\beta }^n=a^n-nb$

• A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
• B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
• C. Assertion is correct but Reason is incorrect
• D. Both Assertion and Reason are incorrect

Answer 1

The correct option is C Assertion is correct but Reason is incorrect

Reason:
As $\alpha ,\beta$ are roots of $x^2-ax+b=0$
Then $\alpha +\beta =a$ and $\alpha \beta =b$
Let $A_n={\alpha }^n+{\beta }^n$
$({\alpha }^{n+1}+{\beta }^{n+1})=(\alpha +\beta )({\alpha }^n+{\beta }^n)-\alpha \beta ({\alpha }^{n-1}+{\beta }^{n-1})\Rightarrow A_{n+1}=(\alpha +\beta )A_n-\alpha \beta A_{n-1}$
So,
$A_2=(\alpha +\beta )(\alpha +\beta )-2\alpha \beta =a^2-2b{A}_3=(\alpha +\beta )(a^2-2b)-\alpha \beta (\alpha +\beta )=a^3-ab$
But from
${\alpha }^n+{\beta }^n=a^n-nb$
$\Rightarrow {\alpha }^3+{\beta }^3=a^3-3b$
Hence reason in wrong
Assertion:
Using reason
$(O{A}_{n+1}{)}^2-a(O{A}_n{)}^2+b(O{A}_{n-1}{)}^2={\alpha }^{n+1}+{\beta }^{n+1}-a({\alpha }^n+{\beta }^n)+b({\alpha }^{n-1}+{\beta }^{n-1})=0$
Hence assertion is correct