Question

i^204+i^206+i^2016=​

Answer 1

Step-by-step explanation:

${i}^{204} + {i}^{206} + {i}^{2016} = > {( {i}^{4} )}^{51} + { ({i}^{2}) }^{103}$

$+ {( {i}^{4})}^{504} = > {1}^{51} + {( - 1)}^{103} + {1}^{504} = >$

$1 - 1 + 1 = > 1$

Hope it helps you.

Answer 2

Answer:

(i'2)'102+(i'2)'103+(i'2)'1008

i'2=-1

(-1)'102+(-1)'103+(-1)'1008

1+(-1)+1

1