Sum of i in C from 0 to 40

Exercise:

Show that , $\sum\limits_{n=0}^{40}{{{i}^{n}}}=1$ where $i\in \mathbb{C}\,,\,\,{{i}^{2}}=-1$

Solution: It’s Clear that ${{i}^{0}}=1$ ( Since ${{\left( {{e}^{i\frac{\pi }{2}}} \right)}^{0}}={{e}^{0}}=1$ )

So $\sum\limits_{n=0}^{40}{{{i}^{n}}={{i}^{0}}+i+{{i}^{2}}+{{i}^{3}}+{{i}^{4}}+...+{{i}^{40}}}$ $=1+i+{{i}^{2}}+{{i}^{3}}+{{i}^{4}}+...+{{i}^{40}}$

$=1+i-1-i+{{\left( {{i}^{2}} \right)}^{2}}+....+{{\left( {{i}^{2}} \right)}^{19}}+{{i}^{39}}+{{\left( {{i}^{2}} \right)}^{20}}=1+i-1-i+1+...-1+i+1=1$