If $ V_n= \alpha^n+\beta^n$ and $\alpha,\beta$ are roots of $x^2+x-1=0$, then $V_n+{V}_{n-3}=2{V}_{n-2}$?

Question

If $ V_n= {\alpha}^n+{\beta}^n$, where ${\alpha}$ and ${\beta}$ are roots of the equation $x^2+x-1=0$. Then prove that $V_n+{V}_{n-3}=2{V}_{n-2}$ (n is whole number).

I have tried to manipulate things on left hand side(using $\alpha + \beta =-1$ and $ \alpha\beta=-1$ from sum and product of roots) but i am unable to get the desired right hand side. Please show me how to manipulate things in order to get the answer. I have tried everything from replacing $\beta$ with $\alpha$ and then squaring/cubing the given equation to get the values of $\alpha^2,\alpha^3,\alpha^4,\alpha^6$. Still not getting it.

Answer 1

Hint: $\;\alpha^2=-\alpha+1\,$, so $\,\alpha^n= -\alpha^{n-1}+\alpha^{n-2}\,$ for $\,n \in \Bbb{Z}\,$, and therefore:

$$\require{cancel} {\alpha}^n+{\alpha}^{n-3} = -\cancel{\alpha^{n-1}}+\alpha^{n-2} +\cancel{\alpha^{n-1}}+\alpha^{n-2} = 2 \alpha^{n-2} $$

Answer 2

.The correct manipulation, in your case, is to note that $$\alpha^n + \beta^n + (\alpha\beta)(\alpha^{n-2} + \beta^{n-2}) = (\alpha + \beta)(\alpha^{n-1} + \beta^{n-1})$$

Since $\alpha+\beta= \alpha \beta = -1$, we get : $V_n - V_{n-2} = -V_{n-1}$.Replacing $n$ by $n-1$,$V_{n-1} - V_{n-3}= -V_{n-2}$ and then substituting for $V_{n-1}$ in the intitial equation gives $V_{n} - V_{n-2} = V_{n-2} - V_{n-3}$, now transpose to get the conclusion, as desired.

---

Note that $\alpha^n + \beta^n$ is therefore an integer, for all $n$. This is not obvious from a first observation.

Answer 3

The LHS is $$V_n+V_{n-3}=\color{red}{\alpha^n}+\color{blue}{\alpha^{n-3}}+\color{green}{\beta^n}+\color{purple}{\beta^{n-3}}$$ and the RHS is $$V_{n-2}=\color{red}{\alpha^{n-2}}+\color{blue}{\alpha^{n-2}}+\color{green}{\beta^{n-2}}+\color{purple}{\beta^{n-2}}.$$ Subtracting (LHS - RHS) gives $$\color{red}{\alpha^{n-2}(\alpha^2-1)}+\color{blue}{\alpha^{n-3}(1-\alpha)}+\color{green}{\beta^{n-2}(\beta^2-1)}+\color{purple}{\beta^{n-3}(1-\beta)}$$ and since $x^2+x-1=0$, we get the expression $$\alpha^{n-2}(-\alpha)+\alpha^{n-3}(\alpha^2)+\beta^{n-2}(-\beta)+\beta^{n-3}(\beta^2)$$ or $$-\alpha^{n-1}+\alpha^{n-1}-\beta^{n-1}+\beta^{n-1}=0$$ hence $$V_n+V_{n-3}-V_{n-2}=0$$ from which you can easily deduce the equality.