If alpha, beta are the roots of x^2 - ax + b | vedclass

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(C) Given that $\alpha, \beta$ are the roots of the quadratic equation $x^2 - ax + b = 0$.By the properties of roots,we have $\alpha + \beta = a$ and

Vedclass · Accepted Answer

$V_{n+1} = aV_n - bV_{n-1}$

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(C) Given that $\alpha, \beta$ are the roots of the quadratic equation $x^2 - ax + b = 0$.By the properties of roots,we have $\alpha + \beta = a$ and $\alpha \beta = b$.We are given $V_n = \alpha^n + \beta^n$.Consider the expression $V_{n+1} = \alpha^{n+1} + \beta^{n+1}$.Since $\alpha$ and $\beta$ are roots,they satisfy the equation $x^2 = ax - b$. Thus,$\alpha^2 = a\alpha - b$ and $\beta^2 = a\beta - b$.Multiplying by $\alpha^{n-1}$ and $\beta^{n-1}$ respectively,we get $\alpha^{n+1} = a\alpha^n - b\alpha^{n-1}$ and $\beta^{n+1} = a\beta^n - b\beta^{n-1}$.Adding these two equations:$\alpha^{n+1} + \beta^{n+1} = a(\alpha^n + \beta^n) - b(\alpha^{n-1} + \beta^{n-1})$.Substituting $V_n$ values,we get $V_{n+1} = aV_n - bV_{n-1}$.

If $\alpha, \beta$ are the roots of $x^2 - ax + b = 0$ and if $\alpha^n + \beta^n = V_n$,then -

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