Question

i+i^2+i^3+....+i^2001 sa se calculeze

Answer 1

i=i

i^2=-1 (stim din definitie)

i^3=i^2 *i=-i

i^4=(i^2)^2=1

i+i^2+i^3+i^4=i-1-i+1=0

la fel e si pt i^5+i^6+i^7+i^8=0

o numim pereche si are 4 termeni

deci 2001/4=500, rest 1

sunt 500 de astfel de perechi care dau 0 si inca un numar, i^2001

i^2001=i* (i^2)^1000=i* (-1)^1000

(-1)^1000=1 =>i^2001=i

deci

i+i^2+i^3+....+i^2001=0+i=i