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.
Best Answer
Hint: $\;\alpha^2=-\alpha+1\,$, so $\,\alpha^n= -\alpha^{n-1}+\alpha^{n-2}\,$ for $\,n \in \Bbb{Z}\,$, and therefore:
$$\require{cancel} {\alpha}^n+{\alpha}^{n-3} = -\cancel{\alpha^{n-1}}+\alpha^{n-2} +\cancel{\alpha^{n-1}}+\alpha^{n-2} = 2 \alpha^{n-2} $$