Question

Evaluate the following:-
$1 + i + {i}^{2} + {i}^{3} + ................ + {i}^{12}$

Answer 1

i^4n = 1
i^4n+1 = i
i^4n+2 = -1
i^4n+3 = -i
where n is a whole no.
so 1+i+i^2+i^3= 1+i-1-i= 0
this is gonna repeat till i^11
so ans is i^12 = 1
solution no 2--------> by sum of GP
a(r^n -1) /(r-1) where n= 13 a= 1
r= i
so sum = 1(i^13 -1) / (i-1)
i^13 = i
thus = i-1/i-1= 1