Hello I am trying to understand the following proof:
$$\begin{align*} (1-x)^n+(1+x)^n &= \sum_{k \geq 0}\binom{n}k (-1)^kx^k + \sum_{k \geq 0} \binom{n}k x^k\\ &= \sum_{k \geq 0}\binom{n}k \left((-1)^k+1\right)x^k\\ &= 2\sum_{k \geq 0}\binom{n}{2k}x^{2k}\\ &< 2\sum_{k \geq 0}\binom{n}{2k}\tag{1}\\ &= 2\cdot 2^{n-1}\tag{2}\\ &= 2^n, \end{align*}$$
But I do not understand why $$\sum_{k \geq 0}\binom{n}k \left((-1)^k+1\right)x^k = 2\sum_{k \geq 0}\binom{n}{2k}x^{2k}.$$ Can someone explain this part? Thank you.
Because $$\frac{1+(-1)^k}{2}=\begin{cases}1 &\text{if $k$ is even}\\0 &\text{if $k$ is odd}\end{cases}$$ we have more generally $$\sum_{k \ge 0} \frac{1+(-1)^k}{2} a_k = \sum_{k \ge 0} a_{2k}.$$ Your identity is the case $a_k=\binom{n}{k}x^k$.