Question 1209762: Let Vₙ = αⁿ + βⁿ where α, β are the roots of the equation x² + x - 1 = 0, then find V₇.
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Let the given equation be $x^2 + x - 1 = 0$.
The roots of this equation are $\alpha$ and $\beta$.
By Vieta's formulas, we have:
$\alpha + \beta = -1$
$\alpha \beta = -1$
Also, since $\alpha$ and $\beta$ are roots of the equation, they satisfy the equation:
$\alpha^2 + \alpha - 1 = 0 \Rightarrow \alpha^2 = 1 - \alpha$
$\beta^2 + \beta - 1 = 0 \Rightarrow \beta^2 = 1 - \beta$
We are given $V_n = \alpha^n + \beta^n$.
We want to find $V_7$.
We can find the first few terms of $V_n$:
$V_0 = \alpha^0 + \beta^0 = 1 + 1 = 2$
$V_1 = \alpha^1 + \beta^1 = \alpha + \beta = -1$
$V_2 = \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = (-1)^2 - 2(-1) = 1 + 2 = 3$
$V_3 = \alpha^3 + \beta^3 = (\alpha + \beta)(\alpha^2 - \alpha\beta + \beta^2) = (\alpha + \beta)((\alpha + \beta)^2 - 3\alpha\beta) = (-1)((-1)^2 - 3(-1)) = (-1)(1 + 3) = -4$
$V_4 = \alpha^4 + \beta^4 = (\alpha^2 + \beta^2)^2 - 2(\alpha\beta)^2 = (3)^2 - 2(-1)^2 = 9 - 2 = 7$
$V_5 = \alpha^5 + \beta^5 = (\alpha^2 + \beta^2)(\alpha^3 + \beta^3) - \alpha^2\beta^2(\alpha + \beta) = (3)(-4) - (-1)^2(-1) = -12 - (-1) = -11$
$V_6 = \alpha^6 + \beta^6 = (\alpha^3 + \beta^3)^2 - 2(\alpha\beta)^3 = (-4)^2 - 2(-1)^3 = 16 - 2(-1) = 16 + 2 = 18$
$V_7 = \alpha^7 + \beta^7 = (\alpha^3 + \beta^3)(\alpha^4 + \beta^4) - \alpha^3\beta^3(\alpha + \beta) = (-4)(7) - (-1)^3(-1) = -28 - 1 = -29$
Alternatively, we can find a recurrence relation for $V_n$.
Since $\alpha$ and $\beta$ are roots of $x^2 + x - 1 = 0$, we have:
$\alpha^2 = 1 - \alpha$
$\beta^2 = 1 - \beta$
Multiplying by $\alpha^{n-2}$ and $\beta^{n-2}$ respectively, we get:
$\alpha^n = \alpha^{n-2} - \alpha^{n-1}$
$\beta^n = \beta^{n-2} - \beta^{n-1}$
Adding the two equations, we have:
$\alpha^n + \beta^n = \alpha^{n-2} + \beta^{n-2} - (\alpha^{n-1} + \beta^{n-1})$
$V_n = V_{n-2} - V_{n-1}$
$V_n = V_{n-2} - V_{n-1}$
$V_0 = 2$
$V_1 = -1$
$V_2 = V_0 - V_1 = 2 - (-1) = 3$
$V_3 = V_1 - V_2 = -1 - 3 = -4$
$V_4 = V_2 - V_3 = 3 - (-4) = 7$
$V_5 = V_3 - V_4 = -4 - 7 = -11$
$V_6 = V_4 - V_5 = 7 - (-11) = 18$
$V_7 = V_5 - V_6 = -11 - 18 = -29$
Final Answer: The final answer is $\boxed{-29}$
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