Answer:
[tex]\sum_{n=0}^9cos(\frac{\pi n}{2})=1[/tex]
[tex] \sum_{k=0}^{N-1}e^{\frac{i2\pi kk}{2}}=0[/tex]
[tex] \sum_{n=0}^\infty (\frac{1}{2})^n cos(\frac{\pi n}{2})=\frac{1}{2}[/tex]
Step-by-step explanation:
[tex] \sum_{n=0}^9cos(\frac{\pi n}{2})=\frac{1}{2}(\sum_{n=0}^9 (e^{\frac{i\pi n}{2}}+ e^{\frac{i\pi n}{2}}))[/tex]
[tex]=\frac{1}{2}(\frac{1-e^{\frac{10i\pi}{2}}}{1-e^{\frac{i\pi}{2}}}+\frac{1-e^{-\frac{10i\pi}{2}}}{1-e^{-\frac{i\pi}{2}}})[/tex]
[tex]=\frac{1}{2}(\frac{1+1}{1-i}+\frac{1+1}{1+i})=1[/tex]
2nd
[tex]\sum_{k=0}^{N-1}e^{\frac{i2\pi kk}{2}}=\frac{1-e^{\frac{i2\pi N}{N}}}{1-e^{\frac{i2\pi}{N}}}[/tex]
[tex]=\frac{1-1}{1-e^{\frac{i2\pi}{N}}}=0[/tex]
3th
[tex] \sum_{n=0}^\infty (\frac{1}{2})^n cos(\frac{\pi n}{2})==\frac{1}{2}(\sum_{n=0}^\infty ((\frac{e^{\frac{i\pi n}{2}}}{2})^n+ (\frac{e^{-\frac{i\pi n}{2}}}{2})^n))[/tex]
[tex]=\frac{1}{2}(\frac{1-0}{1-i}+\frac{1-0}{1+i})=\frac{1}{2}[/tex]
What we use?
We use that
[tex] e^{i\pi n}=cos(\pi n)+i sin(\pi n)[/tex]
and
[tex]\sum_{n=0}^k r^k=\frac{1-r^{k+1}}{1-r}[/tex]
