How to calculate a series with binomial terms invovled I'm studying probability and having trouble in understanding the following calculation

How to get from left to the right on the first line, with the condition that m could only be even numbers? Any hint would be appreciated, thanks~~
 A: We write the same thing, with different symbols. We have 
$$(x+y)^k =\binom{n}{0}x^0y^k+\binom{k}{1}xy^{k-1}+\binom{k}{2}x^2y^{k-2}+\binom{k}{3}x^3y^{n-3}+\cdots.$$
Also,
$$(x-y)^k =\binom{n}{0}x^0y^k-\binom{k}{1}xy^{k-1}+\binom{k}{2}x^2y^{k-2}-\binom{k}{3}x^3y^{k-3}+\cdots.$$
Add, and observe the cancellation. We get
$$(x+y)^k+(x-y)^k=2\binom{k}{0}x^0y^k+2\binom{k}{2}x^2y^{k-2}+2\binom{k}{4}x^4y^{n-4}+\cdots.$$
A: Go from right to left, instead. Note that
\begin{align*}
\left[\left(1 - \frac{1}{n}\right) - \frac{1}{n}\right]^k &= \sum_m {k \choose m} \left(1 - \frac{1}{n}\right)^{k - m} \left(-\frac{1}{n}\right)^{m} \\
&= \sum_m {k \choose m} \left(1 - \frac{1}{n}\right)^{k - m} (-1)^m \left(\frac{1}{n}\right)^m
\end{align*}
Likewise, we can compare this to the other term being added:
$$\left[\left(1 - \frac{1}{n}\right) + \frac{1}{n}\right]^k = \sum_m {k \choose m} \left(1 - \frac{1}{n}\right)^{k - m} \left(\frac{1}{n}\right)^{m}$$
Since $(-1)^m = -1$ if $m$ is odd, the terms corresponding to that $m$ will cancel. When $m$ is even, we'll get two copies of the same term - hence we have to divide by $\frac{1}{2}$.
