<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] You can find the highest degree term in $(x+1)^{2n}-(x-1)^{2n}$, then divide it by $2^n$. Now apply the binomial theorem:
\begin{align}
(x+1)^{2n}&=x^{2n}+2nx^{2n-1}+\text{lower degree terms}\\[4px]
(x-1)^{2n}&=x^{2n}-2nx^{2n-1}+\text{lower degree terms}
\end{align}
Subtracting we get
$$
(x+1)^{2n}-(x-1)^{2n}=4nx^{2n-1}+\text{lower degree terms}
$$
So the required term is
$$
\frac{4n}{2^n}x^{2n-1}
$$
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