<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] You want to express $x^n+b^n$ in terms of their sum, or (for convenience) half-sum $s:=\dfrac{a+b}2$, and geometric mean $p:=\sqrt{ab}$.
We have
$$2as=a^2+ab=a^2+p^2,$$ giving
$$a=s\pm\sqrt{s^2-p^2},b=s\mp\sqrt{s^2-p^2}.$$
Now
$$a^n+b^n=\sum_{k=0}^n\binom nk\left(s^{n-k}(s^2-p^2)^{k/2}+(-1)^ks^{n-k}(s^2-p^2)^{k/2}\right)
\\=2\sum_{j=0}^{2j\le n}\binom n{2j}s^{n-2j}(s^2-p^2)^j.$$
This further expands as
$$2\sum_{j=0}^{2j\le n}\binom n{2j}s^{n-2j}(s^2-p^2)^j
=2\sum_{j=0}^{2j\le n}\binom n{2j}s^{n-2j}\sum_{i=0}^j\binom ji(-1)^is^{2(j-i)}p^{2i}
\\=2s^n\sum_{j=0}^{2j\le n}\sum_{i=0}^j(-1)^i\binom n{2j}\binom ji\left(\frac ps\right)^{2i}.$$
Some more work is needed to regroup the terms by equal $i$.
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