Răspuns:
Explicație pas cu pas:
i · i² · i³ · ........ · iⁿ⁺¹ ; n ∈ N
i = i ; i² = -1 ; i³ = -i ; i⁴ = 1
i·i²·i³·i⁴ = i·(-1)·(-i)·1 = (-i)·(-i) = i² = -1
i·i²·i³·i⁴· .....·i⁸ = (-1)·(-1) = 1
=> i⁴ⁿ = -1 ; n = numar impar
i⁴ⁿ = 1 ; n = numar par
i⁴ⁿ⁺¹ = -i ; n = numar impar
i⁴ⁿ⁺¹ = i ; n = numar par
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i · i² · i³ · ........ · iⁿ⁺¹ = i⁽¹⁺²⁺³⁺........⁺ⁿ⁺¹⁾ = i⁽ⁿ⁺¹⁾⁽ⁿ⁺²⁾/²
1+2+3+.....+n+1 = (n+2)(n+1):2
i⁽ⁿ⁺¹⁾⁽ⁿ⁺²⁾/² = {-1 ; n = 1 ; n = 4k+1 ; k ∈ N
{-i ; n = 2 ; n = 4k+2
{ 1 ; n = 3 ; n = 4k+3
{ i ; n = 4 ; n = 4k+4
