Question 7 - A-Level Maths

Show that $\sum _ { r = 1 } ^ { n } z ^ { 2 r } = \frac { z ^ { 2 n + 1 } - z } { z - z ^ { - 1 } }$, for $z \neq 0,1 , - 1$.
By letting $z = \cos \theta + i \sin \theta$, show that, if $\sin \theta \neq 0$, $$1 + 2 \sum _ { r = 1 } ^ { n } \cos ( 2 r \theta ) = \frac { \sin ( 2 n + 1 ) \theta } { \sin \theta }$$