Question

Evaluate i^30+i^40+i^50+i^60

Answer 1

Given,
${i}^{30} + {i}^{40} + {i}^{50} + {i}^{60} \\ = ( { {i}^{2} })^{15} + ({ {i}^{2} })^{20} + ( { {i}^{2} })^{25} + ( { {i}^{2} })^{30} \\ =( { - 1}^{15}) + ({ - 1}^{20} )+( { - 1}^{25} )+ ({ - 1}^{30} )$
= (-1)+(1)+(-1)+(1)
= 0

We know that,
i² = -1
(-1)^n = 1 [ if n is even ]
(-1)^n = -1 [ if n is odd ]