Question

If S(n)=i^n+i^-n, where i = V-1 and n is an integer, then the total number of distinct values of s(n)is​

Answer 1

Answer:

2 Distinct Values

Step-by-step explanation:

If S(n)=i^n+i^-n, where i = V-1 and n is an integer, then the total number of distinct values of s(n)is​

i   = i   => i⁴ᵃ⁺¹ =  i

i² = -1  => i⁴ᵃ⁺² =  -1

i³ = -i   => i⁴ᵃ⁺³ =  -i

i⁴ = 1    => i⁴ᵃ =  1

i⁻¹ = 1/i  = i/i² = -i   => i⁻⁽⁴ᵃ⁺¹⁾ =  -i

i⁻² = 1/i² = -1   => i⁻⁽⁴ᵃ⁺²⁾ =  -1

i⁻³ = 1/i³  = i   => i⁻⁽⁴ᵃ⁺³⁾ =  i

i⁻⁴ = 1/i⁴  = 1    => i⁻⁽⁴ᵃ⁾ =  1

n Can be   4a  , 4a + a , 4a+2  , 4a + 3

S(n) = S(4a) = i⁴ᵃ +  i⁻⁽⁴ᵃ⁾  = 1 + 1 = 2

S(4a+1) = i⁴ᵃ⁺¹ +  i⁻⁽⁴ᵃ⁺¹⁾  = i - i = 0

S(4a+2) = i⁴ᵃ⁺² +  i⁻⁽⁴ᵃ⁺²⁾  = -1 - 1 = 2

S(4a+3) = i⁴ᵃ⁺³ +  i⁻⁽⁴ᵃ⁺³⁾  = -i + i = 0

S(n)  would be either  0  or 2

so 2 Distinct Values