Summation of series in powers of x with certain combinations as coefficients How can I find the sum: $$\sum_{k=0}^{n} \binom{n-k}{k}x^{k}$$
Edit: The answer to this question is: $$\frac{{(1+\sqrt{1+4x})}^{n+1}-{(1-\sqrt{1+4x})}^{n+1}}{2^{n+1}\sqrt{1+4x}}$$ I don't know how to arrive at this answer.

This answer is similar to this answer and this answer. That is, compute the generating function of the given sequence:
$$
\begin{align}
&\sum_{n=0}^\infty\sum_{k=0}^n\binom{n-k}{k}x^ky^n\\
&=\sum_{k=0}^\infty\sum_{n=k}^\infty\binom{n-k}{k}x^ky^n\tag{1}\\
&=\sum_{k=0}^\infty\sum_{n=0}^\infty\binom{n}{k}x^ky^{n+k}\tag{2}\\
&=\sum_{k=0}^\infty\frac{(xy^2)^k}{(1-y)^{k+1}}\tag{3}\\
&=\frac1{1-y}\frac1{1-\frac{xy^2}{1-y}}\tag{4}\\
&=\frac1{1-y-xy^2}\tag{5}\\
&=\frac1{\left(1-\frac{1-\sqrt{1+4x}}2y\right)\left(1-\frac{1+\sqrt{1+4x}}2y\right)}\tag{6}\\
&=\frac{\frac{-1+\sqrt{1+4x}}{2\sqrt{1+4x}}}{1-\frac{1-\sqrt{1+4x}}2y}+\frac{\frac{1+\sqrt{1+4x}}{2\sqrt{1+4x}}}{1-\frac{1+\sqrt{1+4x}}2y}\tag{7}\\
&=\frac1{\sqrt{1+4x}}\sum_{n=0}^\infty\left[\left(\frac{1+\sqrt{1+4x}}2\right)^{n+1}-\left(\frac{1-\sqrt{1+4x}}2\right)^{n+1}\right]y^n\tag{8}
\end{align}
$$
Explanation:
$(1)$: change order of summation
$(2)$: substitute $n\mapsto n+k$
$(3)$: $\sum\limits_{n\ge k}\binom{n}{k}y^n=\frac{y^k}{(1-y)^{k+1}}$
$(4)$: $\sum\limits_{k\ge 0}x^k=\frac1{1-x}$
$(5)$: simplify
$(6)$: quadratic formula
$(7)$: partial fractions
$(8)$: $\sum\limits_{n\ge 0}x^n=\frac1{1-x}$
Equating coefficients of $y^n$ yields
$$
\sum_{k=0}^n\binom{n-k}{k}x^k
=\frac1{\sqrt{1+4x}}\left[\left(\frac{1+\sqrt{1+4x}}2\right)^{n+1}-\left(\frac{1-\sqrt{1+4x}}2\right)^{n+1}\right]
$$
