Question

Compute i^600+i^599+i^598+....+i+1, where i^2=-1

Quickly please

Answer 1

Write this sum as:


`i^(600)+i^(599)+i^(598)+\ldots+i+1=1+i+i^2+i^3+\ldots+i^(599)+i^(600)`

This is sum of a geometric series with
`n=601` terms, first term
`a=1` and common ratio
`r=i`. So the sum:


`S=a(1-r^n)/(1-r)=1\cdot(1-i^(601))/(1-i)=(1-i\cdot i^(600))/(1-i)=(1-i\cdot(i^2)^(300))/(1-i)=\\\\\\=(1-i\cdot(-1)^(300))/(1-i)= (1-i\cdot1)/(1-i)=(1-i)/(1-i)=\boxed{1}`