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This comes out of a series of transformations, so I'll just get to the main focus here.

Define the functions $$F_n(x)=\frac12\left(x+2+2\sqrt{x+1}\right)^n+\frac12\left(x+2-2\sqrt{x+1}\right)^n. \tag1$$ It's straightforward to obtain $$F_n(x)=\sum_{r=0}^n\binom{2n}{2r}(x+1)^r. \tag2$$

QUESTION 1. On the other hand, how does one obtain (3) by manipulating (1)? $$F_n(x)=\sum_{r=0}^n\frac{n}{n+r}\binom{n+r}{2r}4^rx^{n-r}. \tag3$$

QUESTION 2. As an aside, I'd also be pleased with a combinatorial proof of the sums (2)=(3).

POSTSCRIPT. Question 1 has received ample response and thanks for those who did. I hope someone can address Question 2.

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    $\begingroup$ This doesn't answer your questions, but (2) = (3) is a special case of Pfaff's transformation for the hypergeometric series. More precisely, Pfaff's transformation takes the reversel of (2) to (3). $\endgroup$ – Ira Gessel May 4 at 0:07
  • $\begingroup$ Another useful thing to know is that $x+2+2\sqrt{x+1} = (1+\sqrt{x+1})^2$. $\endgroup$ – Ira Gessel May 19 at 17:40
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Here's a sketch of a derivation of (3) from (1). It's fairly straightforward to compute $$\sum_{n=0}^\infty F_n(x) z^n = \frac{1-2z-xz}{(1-xz)^2 -4z}.$$ If you expand this in powers of $z$ you get (3). (You have to be careful with the constant term, where (3) is undefined.)

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  • $\begingroup$ Thank you for a nice generating function for Question 1. $\endgroup$ – T. Amdeberhan May 5 at 16:30
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The Lucas polynomials $L_n(x,s)=\sum_{j=0}^{\left\lfloor \frac{n}{2} \right\rfloor} \frac{n}{n-j}\binom{n-j}{j}s^jx^{n-2j}$ satisfy the recursion $L_n(x,s)=xL_{n-1}(x,s)+sL_{n-2}(x,s)$ with initial values $L_0(x,s)=2$ and $L_1(x,s)=x.$

Binet’s formula gives $L_n(x,s)= {\left( {\frac{{x + \sqrt {{x^2} + 4s} }}{2}} \right)^n} + {\left( {\frac{{x - \sqrt {{x^2} + 4s} }}{2}} \right)^n}.$

For $2F_n(x)=L_{2n}(2,x)$ this reduces to your formulas.

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  • $\begingroup$ Thank you for a nice recursive approach to Question 1. $\endgroup$ – T. Amdeberhan May 5 at 16:31

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