Is there a closed form for these polynomials?

$P_n(x)=U_n(x/2)$ where $U_n$ is the $n$-th Chebyshev polynomial of the second kind, with (very close) recurrence relationship $U_{n+1}(x)=2x U_n(x)-U_{n-1}$.

A formula that could be of interest for the aim you follow is:

$\dfrac{1}{1-2tx+t^2}=\sum_{k=0}^{\infty}U_k(x)t^k$

(Gradshteyn and Ryzhik formula 8.945)

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(New edit) In connection with the series you are looking for.

1) Your problem is connected with the Darboux-Christoffel kernel.

$K_n(s,t)=\sum_{k=0}^{n}U_k(x)U_k(t)$

Notice that it is not the $P_n$ but the $U_n$.

Let $K_n(x,x)=W_n(x)=\sum_{k=0}^{n}U_k^2(x)$.

One can show that $W_n(x)$ can be expressed as a closed-form formula:

$W_n(x)=U_{n+1}^{\prime}(x)U_{n}(x)-U_{n}^{\prime}(x)U_{n+1}(x)$ (prime for derivative wrt $x$).

(letter $W$ for "Wronskian", a classical tool in ordinary differential equations).

By its definition, $W_n$ is an even function of $x$. It can be proven that the (dominant) degree $2n$ coefficients of $U_{n+1}^{\prime}(x)U_{n}(x)$ and $U_{n}^{\prime}(x)U_{n+1}(x)$ do not cancel. Thus degree($W_n(x)$)=$2n$.

For example, for $n=6$ : $W_6(x)=4-6x^2+30x^4-45x^6+30x^8-9x^{10}+x^{12}$.

I have plotted several curves $y=W_n(x)$ on $[-1,1]$. All of them make small variations above or below a second degree curve, which has its minimum in

$(0,\left \lfloor{\dfrac{n+2}{2}}\right \rfloor)$ and passes through the points $(\pm 1,\left \lfloor{\dfrac{2n+4}{3}}\right \rfloor)$ (where $\left \lfloor a \right \rfloor$ designate the integer part of $a$).

Here is the curve of $W_{30}$:

Thus, experimentaly, one can be fairly confident that for any fixed $x$, the sequence $W_n(x)$ tends to $\infty$ when $n$ tends to $\infty$. But, this is not a proof...

2) In addition, one can read the online article

https://www.ma.utexas.edu/mp_arc/c/08/08-107.pdf

by Barry Simon, a recognized specialist of orthogonal polynomials, especially the $U_n$.

3) Have a look to the similar question (What is the name for the polynomials of the form : $ P_n(x)=2^{-n} \cdot ((x+\sqrt {x^2-4})^n+ (x-\sqrt {x^2-4})^n)$?)

As @JeanMarie states, the recurrence is for $U_n(x/2)$. A cool closed form for these Chebyshevs is

$$U_n(\cos{\theta}) = \frac{\sin{(n+1) \theta}}{\sin{\theta}}$$

so in your case such an expression would look like

$$P_n(x) = \left (1-\frac{x^2}{4} \right )^{-1/2} \sin{\left [(n+1) \arccos{\left ( \frac{x}{2} \right )} \right ]}$$