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

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Question

(C) Since $\alpha$ and $\beta$ are roots of $x^2 - ax + b = 0$,they satisfy the equation:$\alpha^2 - a\alpha + b = 0 \implies \alpha^2 = a\alpha - b$$

Vedclass · Accepted Answer

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

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(C) Since $\alpha$ and $\beta$ are roots of $x^2 - ax + b = 0$,they satisfy the equation:$\alpha^2 - a\alpha + b = 0 \implies \alpha^2 = a\alpha - b$$\beta^2 - a\beta + b = 0 \implies \beta^2 = a\beta - b$Multiplying the first equation by $\alpha^{n-1}$ and the second by $\beta^{n-1}$:$\alpha^{n+1} = a\alpha^n - b\alpha^{n-1}$$\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})$Given $V_n = \alpha^n + \beta^n$,we substitute this into the equation:$V_{n+1} = aV_n - bV_{n-1}$

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

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