If $\alpha$ and $ \beta$ are the roots of the equation $ { x^{ 2 } }-ax+b=0$ and $ { v }_{ n }={ \alpha }^{ n }+{ \beta }^{ n }$, then
- $ { v }_{ n+1 }={ av }_{ n }-{ bv }_{ n }$
- $ { v }_{ n+1 }={ bv }_{ n }-{ av }_{ n-1 }$
- $ { v }_{ n+1 }={ av }_{ n }-{ bv }_{ n-1 }$
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None of the above
Given the quadratic equation x^2 - ax + b = 0, the roots alpha and beta satisfy alpha + beta = a and alpha * beta = b. Since alpha and beta are roots, they satisfy alpha^2 = a*alpha - b and beta^2 = a*beta - b. Multiplying by alpha^(n-1) and beta^(n-1) respectively and adding gives v_{n+1} = a*v_n - b*v_{n-1}.
Since alpha and beta are the roots of x^2 - ax + b = 0, they satisfy the equations alpha^2 = a(alpha) - b and beta^2 = a(beta) - b. We multiply the first relation by alpha^(n-1) and the second by beta^(n-1) to obtain 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 together yields alpha^(n+1) + beta^(n+1) = a(alpha^n + beta^n) - b(alpha^(n-1) + beta^(n-1)). Substituting the given sequence definition v_n = alpha^n + beta^n into this result gives the recurrence relation v_(n+1) = a(v_n) - b(v_(n-1)).