How would I go about finding an expression (preferably closed form) for the sum of $\alpha^n+\beta^n$ in terms of $\alpha + \beta$ and $\alpha\beta$ (where $\alpha$ and $\beta$ are the roots of a quadratic equation. I've done it for the first few cases (up to $n=4$) but I'm unsure of how to formulate a general case.

What I've done so far is say that $\alpha^n + \beta^n = (\alpha + \beta)^n - \displaystyle\sum_{r=1}^{n-1} \binom{n}{r} \alpha^r \beta^{n-r}$ but I don't know what to do from here on.


For $0<r<{n\over 2}$ we write

$${n\choose r}\alpha^r\beta^{n-r}+{n\choose n-r}\alpha^{n-r}\beta^r$$

$$={n\choose r}(\alpha\beta)^r\bigg(\beta^{n-2r}+\alpha^{n-2r}\bigg)$$

Obviously for $n=2k$ we can express

$${n\choose k}\alpha^k\beta^k={n\choose k}(\alpha\beta)^k$$

so there's no problem defining this for $r={n\over 2}$ as well in that case.

Then we proceed inductively. Since


we can express $\alpha^n+\beta^n$ in terms of $\alpha+\beta$ and $\alpha\beta$ for $n\le 2$. If we assume inductively we can do it for all $k\le n$, then this implies there is an expression for $\alpha^n+\beta^n$ in terms of $\alpha+\beta$ and $\alpha\beta$ as well by our expression above.

So you can always write

$$\alpha^n+\beta^n = (\alpha+\beta)^n-\sum_{r=1}^{n/2}{n\choose r}(\alpha\beta)^r\bigg(\alpha^{n-2r}+\beta^{n-2r}\bigg)-(\alpha\beta)^{n/2}\cdot\left\lceil {2-n\over 2}+\left[{n\over 2}\right]\right\rceil$$

the last term being present to subtract off the extra counted ${n\choose {n\over 2}}$ term if it is present.

Note: We check two base cases because the formula requires us to be able to go back two at a time from $n$, so the $n=1$ is the base for odd $n$ and $n=2$ is the base case for even $n$.


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