summation - Sum with irrational powers and binomial coefficients

What is the value of:
$$\sum_{k=1}^{n} {\left(\left(-1\right)^{n-k}{n+1 \choose k+1}\left(\frac{\sqrt5}{1+\sqrt5}\left(3+\sqrt5\right)^k-\frac{\sqrt5}{1-\sqrt5}\left(3-\sqrt5\right)^k-\frac{5}{2}\right)\right)}?$$
I am stuck because of the binomial coefficient there, because without it the sum would just be a bunch of geometric series.

Answer

$\sum_{k=1}^{n} {\left(\left(-1\right)^{n-k}{n+1 \choose k+1}\left(\frac{\sqrt5}{1+\sqrt5}\left(3+\sqrt5\right)^k-\frac{\sqrt5}{1-\sqrt5}\left(3-\sqrt5\right)^k-\frac{5}{2}\right)\right)}?
$

Here's a start.

I'm feeling too tired right now
to do more.

$\begin{array}\\
\sum_{k=1}^{n} {n+1 \choose k+1} x^k
&=\sum_{k=2}^{n+1} {n+1 \choose k} x^{k-1}\\
&=\frac1{x}\sum_{k=2}^{n+1} {n+1 \choose k} x^{k}\\
&=\frac1{x}\left(-1-(n+1)x+\sum_{k=0}^{n+1} {n+1 \choose k} x^{k}\right)\\
&=\frac1{x}\left((1+x)^{n+1}-1-(n+1)x\right)\\
\end{array}

$

Note that
$(3-\sqrt{5})(3+\sqrt{5})
=4
$.

If $x = -(3+\sqrt{5})$,

$\begin{array}\\

\sum_{k=1}^{n} {n+1 \choose k+1}(-1)^k (3+\sqrt{5})^k
&=\frac1{-(3+\sqrt{5})}\left((1-(3+\sqrt{5}))^{n+1}-1+(n+1)(3+\sqrt{5})\right)\\
&=\frac{-(3-\sqrt{5})}{4}\left((-1)^{n+1}(2+\sqrt{5})^{n+1}-1+(n+1)(3+\sqrt{5})\right)\\
\end{array}
$

That's all.