<s>[INST] Closed Form of the Chebyshev Polynomials of the First Kind [Proof Request] $$T_n(x) = \frac{n}{2} \sum_{r=0}^{\lfloor \frac{n}{2} \rfloor} \frac{(-1)^r}{n-r} \binom{n-r}{r} (2x)^{n-2r}$$
Searching on the web yielded no results, and the result is given without proof on OEIS and Wolfram MathWorld
 [/INST] It is defined by $T_n(x)=\frac{u^{n}+u^{-n}}{2}$ where $x=\frac{u+u^{-1}}{2}.$ This implies that $T_{n+1}(x)=2xT_n(x)-T_{n-1}(x)$. So, it is essentially enough to check this recursive relation at $n+1-2r$-the power term:
$$\frac{n+1}{2}\frac{(-1)^r}{n+1-r}{n+1-r\choose r}2^{n+1-2r}=2\frac{n}{2}\frac{(-1)^{r}}{n-r}{n-r\choose r}2^{n-2r}-\frac{n-1}{2}\frac{(-1)^{r-1}}{n-r}{n-r\choose r-1}2^{n+1-2r}$$
which is equivalent to
$$\frac{n+1}{n+1-r}{n+1-r\choose r}=\frac{n}{n-r}{n-r\choose r}+\frac{n-1}{n-r}{n-r\choose r-1}$$
and it is true.
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