<s>[INST] Proof for the general formula for $a^n+b^n$. Based on the following observations. That is

$$a+b = (a+b)^1  \\  a^2+b^2 = (a+b)^2-2ab \\ a^3+b^3 = (a+b)^3-3ab(a+b) \\ a^4+b^4= (a+b)^4-4ab(a+b)^2+2(ab)^2\\ a^5+b^5 
= (a+b)^5 -5ab(a+b)^3+5(ab)^2(a+b)\\\vdots$$

I came to  make the following conjecture as  general formula.

$$ a^n +b^n =\sum_{k=0}^{n-1}(-1)^k \frac{n\Gamma(n-k)}{\Gamma(k+1)\Gamma(n-2k+1)}(a+b)^{n-2k}(ab)^k $$ where $\Gamma(.) $ is gamma function.

I tried up proving the result using binomial theorem $\displaystyle (a+b)^n=\sum_{r=0}^n a^{n-r}b^r$ for positive integers  $a,b$ however, I didn't find  any elegance in the work.  So in the expect of some beautiful proofs,  I wish to share  general formula here.
Thank you
 [/INST] $a$ and $b$ are roots of $x^2=(a+b)x-ab$. Therefore, $a^{n+2}=(a+b)a^{n+1}-(ab)a^n$ and analogously for $b$.
Let $p_n=a^n+b^n$. Then $p_{n+2}=(a+b)p_{n+1}-(ab)p_n$ is a simple recurrence. The initial values are of course $p_0=2$ and $p_1=a+b$.
This recurrence is a special case of Newton's identities.
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