Calculating N-point DFT of a signal based on another signal's DFT
You just need to apply the formula consistently to see what's happening. Noting that $(-1)^n=e^{-j\pi n}$ you get for the DFT of $x[n](-1)^n$

$$\begin{align}\sum_{n=0}^{N-1}x[n](-1)^ne^{-j2\pi nk/N}&amp;=\sum_{n=0}^{N-1}x[n]e^{-j\pi n}e^{-j2\pi nk/N}\\&amp;=\sum_{n=0}^{N-1}x[n]e^{-j2\pi n(k+N/2)/N}\\&amp;=X[k+N/2]\tag{1}\end{align}$$

where $X[k]$ is the DFT of $x[n]$, and where I've assumed that $N$ is even. Consequently, the DFT of $x[n]+(-1)^nx[n]$ is given by

$$\begin{align}\text{DFT}\{x[n]+(-1)^nx[n]\}&amp;=\text{DFT}\{x[n]\}+\text{DFT}\{(-1)^nx[n]\}\\&amp;=X[k]+X[k+N/2]\tag{2}\end{align}$$

where the index $k+N/2$ has to be taken modulo $N$ if $X[k]$ is defined for $k\in[0,N-1]$.