Relation between $S_{n-1}$ , $S_n$ and $S_{n+1}$

Question

We know $ \alpha $ and $ \beta $ are roots of $ax^2+bx+c = 0 $.

also $S_{n-1} = \alpha^{n-1} + \beta^{n-1}$ , $S_{n} = \alpha^{n} + \beta^{n}$ and $S_{n+1} = \alpha^{n+1} + \beta^{n+1}$.

How we can find relation between $S_{n-1}$ , $S_n$ and $S_{n+1}$ ?

Note : We know that relation has $a$ , $b$ and $c$.

Answer 1

Note that $$\alpha^{n} + \beta^{n}=(\alpha + \beta)(\alpha^{n-1} + \beta^{n-1})-\alpha\beta\cdot(\alpha^{n-2} + \beta^{n-2})$$

Now using the relation between the coefficients of a quadratic equation and its roots, we get that

$$S_n=-\frac{b}{a}S_{n-1}-\frac{c}{a}S_{n-2}$$

Or, in other words, we can write that $$S_{n+1}=-\frac{b}{a}S_{n}-\frac{c}{a}S_{n-1}$$

Answer 2

Note that $$S_{n}\cdot(\alpha+\beta)=(\alpha^n+\beta^n)\cdot(\alpha+\beta)=\alpha^{n+1}+\beta^{n+1}+\alpha\beta(\alpha^{n-1}+\beta^{n-1}) =S^{n+1}+\alpha\beta S^{n-1}.$$ Note also that $$\alpha+\beta=-\frac{b}{a}\mbox{ and }\alpha\beta=\frac{c}{a}.$$ Combining all these, we find the required relation.

Answer 3

\begin{align*} \alpha^n+\beta^n &= (\alpha+\beta)(\alpha^{n-1}+\beta^{n-1})- \alpha \beta (\alpha^{n-2}+\beta^{n-2}) \\ &= (\alpha+\beta) S_{n-1}-\alpha \beta S_{n-2} \\ S_{n} &= -\frac{b}{a}S_{n-1}-\frac{c}{a}S_{n-2} \end{align*}

Answer 4

$\alpha$ and $\beta$ satisfy the equation: $ax^2+bx+c=0$ Thus they satisfy the following equation by multiplying with $x^{n-1}$:

$ax^{n+1}+bx^n+cx^{n-1}=0$

Thus:

$a\alpha^{n+1}+b\alpha^n+c\alpha^{n-1}=0$

$a\beta^{n+1}+b\beta^n+c\beta^{n-1}=0$

And by addition:

$aS_{n+1}+bS_n+cS_{n-1}=0$

It generalizes trivially to any polynomial of degree $\geq 2$