Prove that $T_n(x)={}_2F_1\left(-n,n;\tfrac 1 2; \tfrac{1}{2}(1-x)\right) $ Prove that, for  Chebyshev polynomials of the first kind,
\begin{align}
T_n(x) & = \tfrac{n}{2} \sum_{k=0}^{\left \lfloor \frac{n}{2} \right \rfloor}(-1)^k \frac{(n-k-1)!}{k!(n-2k)!}~(2x)^{n-2k} && n>0 \\
& = {}_2F_1\left(-n,n;\tfrac 1 2; \tfrac{1}{2}(1-x)\right) \\
\end{align}
where
$${}_2F_1(a,b;c;z) = \sum_{k=0}^\infty \frac{(a)_k (b)_k}{(c)_k} \frac{z^k}{k!}$$
is the hypergeometric function, and
$$(x)_{n}=x(x-1)(x-2)\cdots(x-n+1).$$
The main difficulty is to understand why $z=\tfrac{1}{2}(1-x)$. In this way, I have
$$\sum_{k=0}^\infty \ldots (1-x)^k$$
and not
$$\sum_{k=0}^\infty \ldots (2x)^k$$
Any suggestions please?

$y=\phantom{}_2 F_1(a,b;c;z)$ is the regular solution of the ODE:
$$z(1-z)y''+\left[c-(a+b+1)z\right]y'-ab y=0 $$
hence $y=\phantom{}_2 F_1(-n,n;1/2;z)$ is the regular solution of the ODE:
$$z(1-z)y''+\left(\frac{1}{2}-z\right)y'+ n^2 y=0 $$
and $y=\phantom{}_2 F_1\left(-n,n;\frac{1}{2};\frac{1-z}{2}\right)$ is the regular solution of the ODE:

$$(1-z^2)\,y'' - z y'+ n^2\,y=0 \tag{1}$$

so to prove our claim we just need to prove that $T_n(z)$ fulfills the same ODE. Since:
$$ T_n(\cos\theta) = \cos(n\theta) $$
by differentiating twice that identity we have:
$$ -\sin(\theta)\, T_n'(\cos\theta) = -n\sin(n\theta),$$
$$ \sin^2(\theta)\, T_n''(\cos\theta) -\cos(\theta)\, T_n'(\cos\theta) = -n^2\cos(n\theta) $$
and $(1)$ just follows from replacing $\cos(\theta)$ with $z$.
