Explicit formula for summation

Question

How can I find an explicit formula for the summation $$\sum_{i=1}^{\left\lfloor\frac{n-1}{2}\right\rfloor+1}\binom{n}{2i-1}\left(\frac{1}{6}\right)^{i}\left(\frac{5}{6}\right)^{n-(2i-1)}$$

Wolfram Alpha comes up with $$-\frac{\left(60+31\sqrt{6}\right)\left[\left(\frac{5}{6}-\frac{1}{\sqrt{6}}\right)^{n}-\left(\frac{5}{6}+\frac{1}{\sqrt{6}}\right)^{n}\right]}{372+120\sqrt{6}},$$ but I have no idea how it came up with it.

Answer 1

The sum can be rewritten as $$\sum_{\substack{0\le k \le n,\\ \color{red}{k \text{ odd}}}} \binom{n}{k}\left(\frac{1}{6}\right)^{\frac{k+1}{2}}\left(\frac{5}{6}\right)^{n-k} =\frac{1}{\sqrt{6}}\sum_{k=0}^n \frac{1-(-1)^k}{2}\binom{n}{k}\left(\frac{\sqrt{6}}{6}\right)^k\left(\frac{5}{6}\right)^{n-k}\\ = \frac{1}{2\sqrt{6}} \left[\left(\frac{5+\sqrt{6}}{6}\right)^n -\left(\frac{5-\sqrt{6}}{6}\right)^n \right] $$

Answer 2

This smells of probability and dice.

$\left(\frac1 6+\frac 5 6\right)^n$ and $\left(\frac1 6-\frac 5 6\right)^n$

Expand them and sum them (or take their differece) then you'll see your $2i-1$ apear. That should be a start.