I am trying to simplify this expression below:
$\Gamma(x,n)=(1+x)^n-(1+x)^{1-n}~\forall~\mathcal{S}:=\{x,n\in\mathbb{R}\lvert x\in(-1,~1),n\geq1\}$
like $\Gamma(x,n)=\eta(.)x$.
So far I am using binomial expansion to simplify this, but the problem is that I am not able to formally define a functional description of $\eta(.)$. I have also given the point where I am stuck:
$\Gamma(x,n)=\left[(2n-1)+\left[\frac{(n-2)+(n+1)}{3!}+\frac{(n-2)(n-3)-(n+1)(n+2)}{4!}x+\frac{(n-2)(n-3)(n-4)+(n+1)(n+2)(n+3)}{5!}x^2\cdots\right]x^2\right]\cdot x$
where,
$\eta(.)=\left[(2n-1)+\left[\frac{(n-2)+(n+1)}{3!}+\frac{(n-2)(n-3)-(n+1)(n+2)}{4!}x+\frac{(n-2)(n-3)(n-4)+(n+1)(n+2)(n+3)}{5!}x^2\cdots\right]x^2\right]$
Could anyone help me to achieve a functional description for the infinite sequence of $\eta(.)$?
For $x \ne 0$ we have:
$\frac{\Gamma(x,n)}{x}=\frac{(1+x)^{2n-1}-1}{x}* \frac{1}{(1+x)^n}=\frac{(1+x)^{2n-1}-1}{1+x-1}* \frac{1}{(1+x)^n}=[(1+(1+x)+(1+x)^2+...+(1+x)^{2n-2}]* \frac{1}{(1+x)^n}$.
Does this help ?