$\sum_{m=0}^{\frac{n}{2}}\binom{n}{2m}(\frac{1}{6})^{2m}(\frac{5}{6})^{n-2m}-\sum_{m=0}^{\frac{n}{2}}\binom{n}{2m+1}(\frac{1}{6})^{2m+1}(\frac{5}{6})^{n-2m-1} = \sum_{k=0}^n\binom{n}{k}(-\frac{1}{6})^k(\frac{5}{6})^{n-k}$
Could someone please explain to me how this subtraction works?
Thanks in advance!
The two series are being interleaved and recast as a single sum.
Notice that in the first sum, $m$ only shows up as the even number $2m$ in the summand. It hits all the terms of the form $$\binom nk \left(\frac16\right)^k \left(\frac56\right)^{n-k}$$ with $k$ even, from $0$ to $n$.
In the second sum, $m$ only shows up as the odd number $2m+1$ in the summand. It hits all terms of the above form with $k$ odd, from $1$ to $n-1$.
The subtraction becomes sign alternation, which we put in the $\left(\frac16\right)^k$ factor since we want it to disappear for $k$ even.