<s>[INST] What does $\Big(\frac{(x+1)^2}{2}\Big)^n-\Big(\frac{(x-1)^2}{2}\Big)^n$ equal to? 
Determine the highest degree term of the polynomial $$\Bigg(\frac{(x+1)^2}{2}\Bigg)^n-\Bigg(\frac{(x-1)^2}{2}\Bigg)^n, \quad n\in\mathbb{N}$$

The answer suggests that the highest degree term is equal to $\dfrac{2nx^{2n-1}}{2^n} + \dfrac{2nx^{2n-1}}{2^n} = \dfrac{4n}{2^n}x^{2n-1}$ But I don't know how to get there. I think it is:
$\begin{split}\Bigg(\dfrac{(x+1)^2}{2}\Bigg)^n-\Bigg(\dfrac{(x-1)^2}{2}\Bigg)^n &= \Bigg(\dfrac{(x+1)^2}{2}-\dfrac{(x-1)^2}{2}\Bigg)\Bigg(\Big(\frac{(x+1)^2}{2}\Big)^{n-1}+\ldots +\Big(\frac{(x-1)^2}{2}\Big)^{n-1}\Bigg) \\ &=2x\Bigg(\Big(\frac{(x+1)^2}{2}\Big)^{n-1}+\ldots +\Big(\frac{(x-1)^2}{2}\Big)^{n-1}\Bigg)\end{split}$ 
 [/INST] Just apply the binomial theorem.
$$\begin{eqnarray*}F_n(x)=\left(\frac{(x+1)^2}{2}\right)^n-\left(\frac{(x-1)^2}{2}\right)^n &=& \frac{1}{2^n}\left[(x+1)^{2n}-(x-1)^{2n}\right]\\&=&\frac{1}{2^n}\sum_{k=0}^{2n}\binom{2n}{k}x^{2n-k}(1^k-(-1)^k) \end{eqnarray*}$$
hence the leading term is $\frac{2\binom{2n}{1}}{2^n}\cdot x^{2n-1}$. You may also notice that:
$$ F_{n+2}(x) = (1+x^2)F_{n+1}(x)-\frac{1}{4}(x^2-1)^2 F_n(x),$$
hence if you prove that the degree of $F_n(x)$ is $2n-1$, the leading coefficient $L_n$ fulfills the recursion:
$$ L_{n+2} = L_{n+1} - \frac{1}{4}L_n $$
hence $L_n = \frac{A+Bn}{2^n}$.
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