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
 A: 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.
A: Starting from the OGF
$$\sum_{n\ge 0} T_n(x) t^n
= \frac{1-tx}{1-2tx+t^2}$$
We extract coefficients $[x^q] [t^n]$ where $0\le q\le n$.
First do the coefficient on $q=0$ which is $[t^n] \frac{1}{1+t^2}.$
Continue with $q\ge 1$
$$[x^q] [t^n] \frac{1-tx}{1-2tx+t^2}
= [x^q] [t^n] \frac{1-tx}{1+t^2} \frac{1}{1-2tx/(1+t^2)}
\\ = [t^n] \frac{1}{1+t^2} \frac{2^q t^q}{(1+t^2)^q}
- [t^n] \frac{1}{1+t^2}
\frac{2^{q-1} t^{q}}{(1+t^2)^{q-1}}
\\ = [t^n] \frac{2^q t^q}{(1+t^2)^{q+1}}
- [t^n] 
\frac{2^{q-1} t^{q}}{(1+t^2)^{q}}.$$
Now we see here that $q$ and $n$ must have the same parity. Supposing
that is true we obtain
$$2^q [t^{n-q}] \frac{1}{(1+t^2)^{q+1}}
- 2^{q-1} [t^{n-q}] \frac{1}{(1+t^2)^q}
\\ = 2^q (-1)^{(n-q)/2} {(n-q)/2+q\choose q}
- 2^{q-1} (-1)^{(n-q)/2} {(n-q)/2+q-1\choose q-1}
\\ = 2^q (-1)^{(n-q)/2} {(n+q)/2\choose q}
- 2^{q-1} (-1)^{(n-q)/2} \frac{q}{(n+q)/2} {(n+q)/2\choose q}
\\ = \frac{n}{n+q} 2^q (-1)^{(n-q)/2} {(n+q)/2\choose q}.$$
Note that this also gives the correct value for $q=0$ so we can merge
that case into our formula. Observe also that we get zero from the
binomial coefficient when $(n+q)/2\lt q$ or $n\lt q.$
Now when $q$ and $n$ have the same parity with $n$ and $q$ non-negative
then $n-q=2r$ where $0\le r\le \lfloor n/2 \rfloor.$ We have $q=n-2r$
and obtain summing over $r$
$$\sum_{r=0}^{\lfloor n/2 \rfloor} x^{n-2r}
\frac{n}{2n-2r} 2^{n-2r} (-1)^r {n-r\choose n-2r}
\\ = \frac{n}{2} \sum_{r=0}^{\lfloor n/2 \rfloor} (2x)^{n-2r}
\frac{(-1)^r}{n-r} {n-r\choose r}.$$
This is the claim.
