Question

Value of (iⁿ+iⁿ⁺¹+iⁿ⁺²+iⁿ⁺³) is

Answer 1

Answer:

- 2

Step-by-step explanation:

i = √(-1)

1). If (n) is odd number then

(n+1) and (n+3) are even

i^n - 1 + (i^n * i^2) - 1 = i^n - i^n - 2 = - 2

2). If "n" is even number then (n+1) and (n+3) are odd

- 1 + i - 1 - i = - 2

Answer 2

Answer:

The answer is $i^{n}(0)=0$.

Step-by-step explanation:

A complex number exists as an element of a number system that includes the real numbers and a specific element suggested i, named the imaginary unit and satisfying the equation i2 = −1. Moreover, every complex number can be represented in the form a + bi, where a and b are real numbers.

Given,

(iⁿ+iⁿ⁺¹+iⁿ⁺²+iⁿ⁺³).

To find,

The Value of (iⁿ+iⁿ⁺¹+iⁿ⁺²+iⁿ⁺³).

Step 1

Given,

$i^{2}=-1, i^{3}=i^{2} \cdot i=-i$

$\Longrightarrow i^{n}\left(1+i+i^{2}+i^{3}\right)$

Simplifying,

$\Longrightarrow i^{n}(1+i-1-i)$

We get,

$i^{n}(0)=0$.

#SPJ2