Problems computing the DFT of finite length sequence

Question

I am having trouble finding the same answer as the solution manual for this sequence.

The problem asks to compute the DFT of $$ x[n] = \begin{cases} 1 & \text{for even } n \in \{0\ldots N-1\} \\ 0 & \text{for odd } n \in \{0\ldots N-1\} \end{cases} $$

Setting up the DFT equation I agree with the solution manual $$ X[k]=\sum_0^{N-1}x[n]W^{kn}_{N}=\sum_0^{\frac{N}{2}-1}W^{k2n}_{N}=\sum_0^{\frac{N}{2}-1}e^{\frac{-j2\pi k(2n)}{N}} $$

However when apply the closed form geometric series formula $\sum_0^Na^k=\frac{1-a^{N+1}}{1-a}$ I get $$ \frac{1-e^{-j2\pi k}}{1-e^{\frac{-j2\pi k}{N}}} $$

while the solution manual says its
$$\frac{1-e^{-j2\pi k}}{1-e^{\frac{-j\pi k}{N}}}$$

Note their lower exponent does not have a 2. Could someone explain where and how the 2 goes away? Or does someone agree that the provided answer is incorrect?

This is question 8.5c from Oppenheim's Discrete-Time Signal Processing 3e.

Answer 1

Assuming $N$ even :

$$ \begin{align} X[k] &= \sum_{n=0}^{N-1} x[n] ~ W_N^{kn} ~~~,~~~k=0,1,...,N-1\\ \\ &= \sum_{n=0}^{N/2-1} 1 ~ W_N^{k ~2n} \\ \\ &= \sum_{n=0}^{N/2-1} e^{-j 4\pi kn/N} \\ \\ & = \frac{ 1 - e^{-j \frac{4\pi k}{N}N/2 }}{ 1 - e^{-j 4\pi k/N}} \\ \\ & = \frac{ 1 - e^{-j 2\pi k}}{ 1 - e^{-j 4\pi k/N}}~~~,~~~k=0,1,...,N-1\\ \end{align} $$

The result equals: $$ X[k] = \begin{cases}{ ~~~ N/2 ~~~ ,~~~ k = 0, N/2 \\ ~~~~~~ 0 ~~~~~ ~ ,~ ~\text{otherwise} } \end{cases} $$

Answer 2

Figured I'd post this since I wrote it anyway, just a confirmation of Fat32's answer.

Letting $N' = \frac{N}{2} - 1$ we have $\sum_{0}^{\frac{N}{2} - 1} e^{\frac{-j2\pi k (2n)}{N}} = \sum_{0}^{N'} e^{\frac{-j2\pi k n}{(N'+1)}}$

Then plugging in the geometric sum formula:

$=\frac{1 - e^{\frac{-j2\pi k(N'+1)}{(N'+1)}}}{1-e^{\frac{-j2\pi k}{(N'+1)}}}=\frac{1 - e^{\frac{-j2\pi k}{1}}}{1-e^{\frac{-j2\pi k}{(N'+1)}}}$

and finally reverting back to our original variable $N=2(N'+1)$ gives $=\frac{1 - e^{\frac{-j2\pi k}{1}}}{1-e^{\frac{-j4\pi k}{N}}}$