An expression involving trig functions and binomial coefficients Prove that
\begin{eqnarray}\sum_{i=0}^k5^{k-i-1}\binom{k}{i}\left(\frac{(37+62\cos\frac{2\pi}{5}+56\cos\frac{4\pi}{5})^i}{\left(2\sin\frac{\pi}{5}\right)^{2(k-i-1)}}+\frac{(37+62\cos\frac{4\pi}{5}+56\cos\frac{8\pi}{5})^i}{\left(2\sin\frac{2\pi}{5}\right)^{2(k-i-1)}}\right)\end{eqnarray}
is an integer. Can we further simplify the above expression? Thanks.
 A: Using the same approach as @drhab in his answer, I arrived to a slightly different result
$$\color{blue}{t_k=\frac{b}{5}\left(a+\frac 5b\right)^{k}+\frac{d}{5}\left(c+\frac5d\right)^{k}}$$ Now, using the values of the trigonometric terms, we have
$$a=\frac{3}{2} \left(5+\sqrt{5}\right)\qquad b=\frac{1}{2} \left(5-\sqrt{5}\right)\qquad c=\frac{3}{2} \left(5-\sqrt{5}\right)\qquad d=\frac{1}{2} \left(5+\sqrt{5}\right)$$
$$a+\frac 5b=2 \left(5+\sqrt{5}\right)\qquad c+\frac5d=2 \left(5-\sqrt{5}\right)$$  leading the the sequence
$$\{1,8,80,960,12800,179200,2560000,36864000,532480000,7700480000,111411200000\}$$
We could make $t_k$ more fancy in terms of $\phi$, the golden ratio.
A: You could start with something like:
$$\sum_{i=0}^{k}5^{k-i-1}\binom{k}{i}\left[\frac{a^{i}}{b^{k-i-1}}+\frac{c^{i}}{d^{k-i-1}}\right]=\frac{b}{5}\sum_{i=0}^{k}\binom{k}{i}a^{i}\left(\frac{5}{b}\right)^{k-i}+\frac{d}{5}\sum_{i=0}^{k}\binom{k}{i}c^{i}\left(\frac{5}{d}\right)^{k-i}=$$$$\frac{b}{5}\left(a+\frac{5}{b}\right)^{k}+\frac{d}{5}\left(c+\frac{5}{d}\right)^{k}$$
This to get rid of the sum.
I hope it helps, but too risky to give any guarantees :-).
