1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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

$$\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta$$

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 $n$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.