After inspiring by Cardano method for $3th$ degree equations, I noticed that it may be possible to express $(x+y)^n$ with only these terms: $x^n,y^n, xy,x+y $
Let's see some examples:
$(x+y)^1=x^1+y^1$
$(x+y)^2=x^2+y^2+2*(xy)$
$(x+y)^3=x^3+y^3+3*(xy) *(x+y)$ $-$ (Cardano method uses this equality)
$(x+y)^4=x^4+y^4+4*(xy) *(x+y)^2-2*(xy)^2$
$(x+y)^5=x^5+y^5+5*(xy)*(x+y)^3-5*(xy)^2*(x+y)$
$(x+y)^6=x^6+y^6+6*(xy)*(x+y)^4-9*(xy)^2*(x+y)^2+2*(xy)^3$
$(x+y)^7=x^7+y^7+7*(xy)*(x+y)^5-14*(xy)^2*(x+y)^3+7*(xy)^3*(x+y)$
...
So I decided to generalize these formulas and came up with this formula:
$$(x+y)^n=x^n+y^n+\sum_{i=0; 2i<=n-2}^{} {(-1)^i*H_n^i*(xy)^{1+i}*(x+y)^{n-2-2i}}$$
Here $H_n^i$ are positive numbers depending on $n$ and $i$
I only found some small values of it such that:
$$H_n^0=n$$
$$H_n^1=\frac{n*(n-3)}{2}$$
I would like to ask:
- Can we prove that it is true for all $n$ and find formula for $H_n^i$
- Any similar formula for 3 or more terms like : $(x+y+z)^n$
