Hannah02
10

# How do I prove that 1+r+r^2+r^3+r^4+r^5=(1-r^6)/(1-r) ?

let's say $A=1+r+r^2+r^3+r^4+r^5$ let's multiple all member by r $r*A=r+r^2+r^3+r^4+r^5+r^6$ Now we subtract $A-r*A=(1+r+r^2+r^3+r^4+r^5)-(r+r^2+r^3+r^4+r^5+r^6)$ $A-r*A=1+r+r^2+r^3+r^4+r^5-r-r^2-r^3-r^4-r^5-r^6$ $A(1-r)=1-r^6$ therefore... $\boxed{\boxed{\boxed{\therefore~A=\frac{(1-r^6)}{(1-r)}}}}$