New answers tagged self-study
1
HINT:
$$\sum_{n=5}^9e^{-j\pi kn/5}=\sum_{n=0}^4e^{-j\pi k(n+5)/5}=\sum_{n=0}^4e^{-j\pi kn/5}e^{-j\pi k}=(-1)^k\sum_{n=0}^4e^{-j\pi kn/5}$$
I'm sure you can take it from here.
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