Is there a closed form of $\sum\limits_{r=0}^n \binom{n-r}{r}x^r$?

Question

$\sum\limits_{r=0}^n \binom{n-r}{r}x^r=\sum\limits_{r=0}^{\lfloor n/2\rfloor} \binom{n-r}{r}x^r$

I need its closed form for a probability problem. I know about the case where $x=1$. It's the sum of the shallow diagonals of the Pascal's triangle and it equals $(n+1)^{\texttt{th}}$ number in the Fibonacci sequence.

$\sum\limits_{r=0}^{\lfloor n/2\rfloor} \binom{n-r}{r}=\frac{\phi^{n+1}-\varphi^{n+1}}{\sqrt 5}$

Here, $\phi$ and $\varphi$ are the positive and negative roots of the equation $x^2-x-1=0$ respectively.

Answer 1

Working after @Greg Martin's comment $$t=\sqrt {4x+1}~\implies~ U_n=\frac{ (1+t)^{n+1}-2 (1-t)^{n+1}}{2^{n+1}\,\,t}$$

Answer 2

I just want to share this knowledge. There is a computer algebra system called Sage, which can do a lot of crazy maths. In this case, it is able to find the closed form found by Claude Leibovici above directly. You can try it out online here! Try this code:

x, n, r = var('x, n, r')
print(sum(binomial(n - r, r) * x^r, r, 0, n)

Output:

1/2*((4*x + 1)*(1/2*sqrt(4*x + 1) + 1/2)^n + (4*x + 1)*(-1/2*sqrt(4*x + 1) + 1/2)^n + sqrt(4*x + 1)*((1/2*sqrt(4*x + 1) + 1/2)^n - (-1/2*sqrt(4*x + 1) + 1/2)^n))/(4*x + 1)

This is quite unreadable for me :P So here's another trick:

print(latex(sum(binomial(n-r,r)*x^r,r,0,n)))

Output (directly rendered with latex)

$$ \frac{{\left(4 \, x + 1\right)} {\left(\frac{1}{2} \, \sqrt{4 \, x + 1} + \frac{1}{2}\right)}^{n} + {\left(4 \, x + 1\right)} {\left(-\frac{1}{2} \, \sqrt{4 \, x + 1} + \frac{1}{2}\right)}^{n} + \sqrt{4 \, x + 1} {\left({\left(\frac{1}{2} \, \sqrt{4 \, x + 1} + \frac{1}{2}\right)}^{n} - {\left(-\frac{1}{2} \, \sqrt{4 \, x + 1} + \frac{1}{2}\right)}^{n}\right)}}{2 \, {\left(4 \, x + 1\right)}} $$

Not the most simplified form, but here you go :)