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.

Introduce the generating function
$$f(z) = \sum_{n\ge 0} z^n \sum_{k=0}^n {n-k\choose k} x^k.$$
This becomes
$$f(z) = \sum_{k\ge 0} x^k 
\sum_{n\ge k} z^n {n-k\choose k}
\\ = \sum_{k\ge 0} x^k 
\sum_{n\ge 0} z^{n+k} {n\choose k}
= \sum_{k\ge 0} x^k 
\sum_{n\ge k} z^{n+k} {n\choose k}
\\ = \sum_{k\ge 0} x^k 
\sum_{n\ge 0} z^{n+2k} {n+k\choose k}
= \sum_{k\ge 0} x^k z^{2k}
\sum_{n\ge 0} z^{n} {n+k\choose k}
\\ = \sum_{k\ge 0} x^k z^{2k}
\frac{1}{(1-z)^{k+1}}
= \frac{1}{1-z} \frac{1}{1-xz^2/(1-z)}
\\ = \frac{1}{1-z-xz^2}
= -\frac{1/x}{z^2+z/x-1/x}.$$
Solving $z^2+z/x-1/x$ we obtain $$\rho_{1,2}
= \frac{-1 \pm\sqrt{1+4x}}{2x}$$
and we have that
$$f(z) = -\frac{1}{x}\frac{1}{(z-\rho_1)(z-\rho_2)}
\\ = -\frac{1}{x}
\frac{1}{\rho_1-\rho_2}
\left(\frac{1}{z-\rho_1}-\frac{1}{z-\rho_2}\right)
\\ = -\frac{1}{x}
\frac{1}{\rho_1-\rho_2}
\left(\frac{1}{\rho_1}\frac{1}{z/\rho_1-1}
-\frac{1}{\rho_2}\frac{1}{z/\rho_2-1}\right).$$
Extracting coefficients from this we obtain
$$-\frac{1}{x}
\frac{x}{\sqrt{1+4x}}
\left(-\frac{1}{\rho_1^{n+1}}
+\frac{1}{\rho_2^{n+1}}\right).$$
Now since $\rho_1\rho_2 = -1/x$ this becomes
$$-\frac{1}{\sqrt{1+4x}}
\left(-(-1)^{n+1} x^{n+1} \rho_2^{n+1}
+(-1)^{n+1} x^{n+1} \rho_1^{n+1}\right).$$
Observe that
$$(-1)^{n+1} x^{n+1} \rho_{1,2}^{n+1}
= \frac{(1\mp\sqrt{1+4x})^{n+1}}{2^{n+1}}$$
so this finally becomes
$$-\frac{1}{\sqrt{1+4x}}
\frac{(1-\sqrt{1+4x})^{n+1}-(1+\sqrt{1+4x})^{n+1}}
{2^{n+1}}
\\ = \frac{1}{\sqrt{1+4x}}
\frac{(1+\sqrt{1+4x})^{n+1}-(1-\sqrt{1+4x})^{n+1}}
{2^{n+1}}.$$

Alternate derivation of the generating function.
Introduce
$${n-k\choose k}
= \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{n-k}}{z^{k+1}} dz.$$
This yields for the sum
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{n}}{z} 
\sum_{k=0}^n \frac{x^k}{(1+z)^k z^k} \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{n}}{z}
\frac{x^{n+1}/(1+z)^{n+1}/z^{n+1}-1}
{x/(1+z)/z-1} \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} (1+z)^{n+1}
\frac{x^{n+1}/(1+z)^{n+1}/z^{n+1}-1}
{x-z-z^2} \; dz.$$
We omit the part that does not contribute to get
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon} (1+z)^{n+1}
\frac{x^{n+1}/(1+z)^{n+1}/z^{n+1}}
{x-z-z^2} \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n+1}}
\frac{x^{n+1}}{x-z-z^2} \; dz.$$
This is
$$[z^n] \frac{x^{n+1}}{x-z-z^2}
= x [z^n] x^{n} \frac{1}{x-z-z^2} 
\\ = x [z^n] \frac{1}{x-zx-z^2x^2} 
= [z^n] \frac{1}{1-z-z^2x}.$$
We have the same generating function as above, QED.
